Continuity

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x), we can say that if we plot the coordinates (x, f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means:–

Limiting the value of the left neighbourhood of x = c also called left–hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds, function is continuous else it is discontinuous.

2. The idea of modulus function |x |: You can think this function as a machine in which you can give it any real no. as an input and it returns its absolute value i.e. if positive is entered it returns the same no and if negative is entered it returns the corresponding positive no.

Eg:– |2| = 2 ; |–2| = –(–2) = 2

Similarly, we can define it for variable x, if x ≥ 0 |x| = x

If x < 0 |x| = (–x)

∴

Now we are ready to solve the question –

We need to check the continuity at the origin (0,0) i.e. we will check it at x=0.

So, we need to see whether at x=0,

IF, LHL = RHL = f(0)

i.e.

For this question c = 0

f(x) can be rewritten using the concept of modulus function as :

…… Equation 1

NOTE :

Now we have three different expressions for different conditions of x.

LHL = = = using eqn 1

RHL = using eqn 1

LHL ≠ RHL so we even don’t need to check for f(0)

∴ We can easily say that f(x) is discontinuous at the origin.

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x), we can say that if we plot the coordinates (x, f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means:–

Limiting the value of the left neighborhood of x = c also called left–hand limit LHL must be equal to limiting value of right neighborhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

Let’s solve :

To prove function is continuous at x=3 we need to show LHL = RHL = f(c) As continuity is to be checked at x = 3, therefore c=3. (in equation 1)

……Eqn 2

∴ f(3) = 5 using eqn 2

LHL =

Using equation 2 –

RHL =

Clearly, LHL = RHL = f(3) = 5

∴ f(x) is continuous at x=3

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x), we can say that if we plot the coordinates (x, f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means:–

Limiting the value of the left neighborhood of x = c also called left–hand limit LHL must be equal to limiting value of right neighborhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds, function is continuous else it is discontinuous.

Let’s solve :

To prove function is continuous at x=3, we need to show LHL = RHL = f(c) As continuity is to be checked at x = 3 therefore c=3 (in equation 1)

…… eqn 2

From eqn 2 :

f(3) = 6

LHL =

Using equation 2 –

RHL =

Clearly, LHL = RHL = f(3) = 6

∴ f(x) is continuous at x=3

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x), we can say that if we plot the coordinates (x, f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds, function is continuous else it is discontinuous.

Let’s solve :

To check whether function is continuous at x=3 we need to check whether LHL = RHL = f(c)

As continuity is to be checked at x = 1 therefore c=1 (in equation 1)

As, ………eqn 2

From eqn 2 :

f(1) = 2

LHL =

Using equation 2 –

RHL =

Clearly, LHL = RHL = f(1) = 2

∴ f(x) is continuous at x=1

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x), we can say that if we plot the coordinates (x, f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means:–

Limiting the value of the left neighbourhood of x = c also called left–hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds, function is continuous else it is discontinuous.

2. The idea of sandwich theorem – This theorem also known as squeeze theorem that you may have encountered in your class 11 in limits chapter suggests that

If I be an interval having a point as a limit point. Let g, f, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have{\displaystyle \lim _{x→ a}g(x)=\lim _{x→ a}h(x)=L.}{\displaystyle \lim _{x→ a}f(x)=L.}

g(x) ≤ f(x) ≤ h(x) and also

Then , We say that f(x) is squeezed between g(x) and h(x) or you can assume it like sandwich.

∴ equation 2

NOTE : denominator in the above limit should be exactly same as that of content in sine function

Eg :

Let’s solve :

To check whether function is continuous at x=0 we need to check whether LHL = RHL = f(c)

As continuity is to be checked at x = 0 therefore c=0.

As, Equation 3

Clearly, f(0) = 1 using eqn 3

LHL =

Using equation 3 –

[∵ sin(–θ) = – (sin θ)]

To find its limit we need to think of Sandwich theorem as the form looks similar but term in denominator is not exactly same as that in the content of sine function.

Ok no problem lets bring it there but if we put 3 in denominator we need to put a 3 in numerator too so that both can be cancelled.

Let’s do it:

∴ LHL

RHL =

Using equation 3 –

Again using sandwich theorem as we used while finding LHL

∴ RHL

Thus,

LHL = RHL = 3

But, f(0) = 1 by definition of f(x) i.e. equation 3

∴ LHL = RHL ≠ f(0)

∴ condition for function to be continuous is not satisfied

∴ f(x) is discontinuous at x = 0

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

Let’s solve :

To check whether function is continuous at x=0 we need to check whether LHL = RHL = f(c)

As continuity is to be checked at x = 0 therefore c=0.

As, …… say it equation 2

Clearly, f(0) = 1 using eqn 3

LHL =

Using equation 3 –

[∵ h is very small , is very large ‘–ve‘]

RHL =

Using equation 3 –

[∵ h is very small , is very large ‘+ve‘]

Clearly, LHL ≠ RHL ≠ f(0)

∴ We can easily say that f(x) is discontinuous at origin.

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

2. Idea of sandwich theorem – This theorem also known as squeeze theorem that you may have encountered in your class 11 in limits chapter suggests that

If I be an interval having the point a as a limit point. Let g, f, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have{\displaystyle \lim _{x→ a}g(x)=\lim _{x→ a}h(x)=L.}{\displaystyle \lim _{x→ a}f(x)=L.}

g(x) ≤ f(x) ≤ h(x) and also

Then , We say that f(x) is squeezed between g(x) and h(x) or you can assume it like sandwich.

∴ equation 2

NOTE : denominator in the above limit should be exactly same as that of content in sine function

Eg :

Let’s solve :

To check whether function is continuous at x=0 we need to check whether LHL = RHL = f(c)

As continuity is to be checked at x = 0 therefore c=0.

As, …… Equation 3

Clearly, f(0) = 1 using eqn 3

LHL =

Using equation 3 –

[∵ cos(–θ) = (cosθ)]

Whenever you see trigonometric term in numerator and corresponding or similar content of term in numerator try to apply sandwich theorem in finding apply if you are unable to do so think of alternative.

Here we have cos term but to apply the theorem we need to have sin term in numerator.

Can we bring it ?

Yes, we know that :

(1 – cos θ) = 2/2)

We have,

LHL = 2

To find its limit we need to think of Sandwich theorem as the form looks similar but term in denominator is not exactly same as that in the content of sine function.

Ok no problem, let’s bring it there but if we multiply 1/2 in denominator we need to multiply 1/2 in numerator too so that both can be cancelled.

Let’s do it:

∴ LHL =

= =

RHL =

Using equation 3 –

It is not taking or any other indeterminate form.

So, RHL =

Thus,

LHL ≠ RHL

∴ LHL ≠ RHL ≠ f(0)

∴ condition for function to be continuous is not satisfied

∴ f(x) is discontinuous at x = 0

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

..Equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

2. Idea of modulus function |x| : You can think this function as a machine in which you can give it any real no. as an input it returns it absolute value i.e if positive is entered it returns the same no and if negative is entered it returns the corresponding positive no.

Eg:– |2| = 2 ; |–2| = –(–2) = 2

Similarly, we can define it for variable x, if x ≥ 0 |x| = x

If x < 0 |x| = (–x)

∴

Now we are ready to solve the question –

We need to check the continuity at origin (0,0) i.e we will check it at x=0.

So, we need to see whether at x=0,

LHL = RHL = f(0)

i.e.

For this question c = 0

f(x) can be rewritten using the concept of modulus function as :

…… Equation 2

Now we have three different expressions for different conditions of x.

LHL = = = using eqn 2

RHL = using eqn 2

f(0) = 2

Clearly, LHL = RHL ≠ f(0)

∴ We can easily say that f(x) is discontinuous at origin.

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

……Equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

2. Idea of modulus function |x| : You can think this function as a machine in which you can give it any real no. as an input it returns it absolute value i.e if positive is entered it returns the same no and if negative is entered it returns the corresponding positive no.

Eg:– |2| = 2 ; |–2| = –(–2) = 2

Similarly, we can define it for variable x, if x ≥ 0 |x| = x

If x < 0 |x| = (–x)

∴

Now we are ready to solve the question –

We need to check the continuity at x = a

So, we need to see whether at x=a,

LHL = RHL = f(a)

i.e.

For this question c = a

We have:

Using the concept of modulus function we can redefine the above equation as given below :

f(x) = ……Equation 2

Clearly,

f(a) = 1

LHL = = using eqn 2

RHL = using eqn 2

Clearly, LHL ≠ RHL

∴ We can easily say that f(x) is discontinuous at origin.

Idea to solve problem :

Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

……Equation 1

where h is a very small positive no (can assume h = 0.00000000001 or even smaller )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

Let’s Solve the problem now:

we need to check continuity at x = 0 for given function

……equation 2

∴ we need to check RHL ,LHL and value of function at x = 0

Clearly,

f(0) = 0 [from equation 2]

LHL = [∵ cos (–θ) = cos θ]

RHL = [from eqn 2]

Why

Since whatever is value of h, cos(1/h) is goimg to range from –1 to 1

As h→ 0 i.e. h is approximately 0

∴ 0*some finite quantity is equal to 0.

Clearly, LHL = RHL = f(0)

∴ f(x) is continuous at x = 0

In this problem we need to check continuity at x = 0

Given function is

at x = 0 …… Equation 2

∴ we need to check LHL, RHL and value of function at x = 0 (for idea and meaning of continuity refer to Q10(i) )

Clearly,

f(0) = 0 [from equation 2]

LHL = [∵ sin(–θ) = sinθ]

RHL = [ using eqn 2]

Why

Since whatever is value of h, sin(1/h) is going to range from –1 to 1

As h→ 0 i.e. h is approximately 0

∴ 0*some finite quantity is equal to 0.

Clearly, LHL = RHL = f(0)

∴ f(x) is continuous at x = 0

In this problem we need to check continuity at x = a

Given function is

at x = a …… Equation 2

∴ we need to check LHL, RHL and value of function at x = a (for idea and meaning of continuity refer to Q10(i)

Clearly,

f(a) = 0 [from equation 2]

LHL = [∵ sin(–θ) = sinθ]

RHL = [from eqn 2]

Why

Since whatever is value of h, sin(1/h) is going to range from –1 to 1

As h→ 0 ,i.e. approximately 0

∴ 0*some finite quantity is equal to 0.

Clearly, LHL = RHL = f(a)

∴ f(x) is continuous at x = 0

In this problem we need to check continuity at x = 0

Given function is

at x = 0

∴ we need to check LHL, RHL and value of function at x = 0 (for idea and meaning of continuity refer to Q10(i))

1. NOTE : Idea of logarithmic limit and exponential limit –

……equation 1

…… equation 2

You must have read such limits in class 11. You can verify these by expanding log(1+x) and e^x in its taylor form.

Numerator and denominator conditions also hold for this limit like sandwich theorem.

E.g :

But ,

Now we are ready to solve the problem.

Given function is

at x = 0 …… Equation 3

Clearly,

f(0) = 7 [from equation 2]

LHL = [ putting x = –h in equation 3]

=

Using logarithmic and exponential limit as explained above, we have:

LHL =

RHL = [ putting x = h in equation 3]

=

Using logarithmic and exponential limit as explained above, we have:

RHL =

Thus, LHL = RHL ≠ f(0)

∴ f(x) is discontinuous at x = 0

In this problem we need to check continuity at x = a

Given function is

at x = 1 …… Equation 2

∴ we need to check LHL, RHL and value of function at x = a (for idea and meaning of continuity refer to Q10(i) )

Clearly,

f(1) = n–1 [from equation 2]

LHL =

Using binomial theorem –

(1–h)ⁿ = 1 – nh + h² – …….

LHL =

Putting h=0 we get,

LHL = n

RHL =

Using binomial expansion as used above we get the following expression

Similarly,

RHL =

Putting h=0 we get,

RHL = n

Thus RHL = LHL ≠ f(1)

∴ f(x) is discontinuous at x=1

In this problem we need to check continuity at x = 1

Given function is

at x = 1 …… Equation 2

∴ we need to check LHL, RHL and value of function at x = 1 (for approaching idea and meaning of continuity refer to Q10(i))

Clearly,

f(1) = 2 [ from equation 2]

LHL =

Since h is positive no which is very close to 0

∴ (h–2) is negative and hence h(h–2) is also negative.

∴ |h(h–2)| = –h(h–2)

∴LHL =

RHL =

Since h is a positive no which is very close to 0

∴ (h+2) is positive and hence h(h–2) is also positive.

∴ |h(h+2)| = h(h+2)

∴ RHL =

Clearly, LHL ≠ RHL

∴ f(x) is discontinuous at x=1

In this problem we need to check continuity at x = 1

Given function is

at x = 0 …… Equation 2

∴ we need to check LHL, RHL and value of function at x =0 (for approaching idea and meaning of continuity refer to Q10(i))

Clearly,

f(0) = 0 [ from equation 2]

LHL = [by putting x = –h in eqn 2]

=

RHL = [by putting x = h in eqn 2]

=

Clearly, LHL ≠ RHL ≠ f(0)

∴ f(x) is discontinuous at x=0

In this problem we need to check continuity at x = a

Given function is

at x = a …… Equation 2

∴ we need to check LHL, RHL and value of function at x = a (for idea and meaning of continuity refer to Q10(i))

Clearly,

f(a) = 0 [from equation 2]

LHL = [putting x = (a–h) in eqn 2]

=

RHL =

=

Why

Since whatever is value of h, sin(1/h) is going to range from –1 to 1

As h→ 0 ,i.e. approximately 0

∴ 0*some finite quantity is equal to 0.

Clearly, LHL = RHL = f(a)

∴ f(x) is continuous at x = 0

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

……Equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

In this problem we need to check continuity at x = 1

Given function is

…… Equation 2

∴ we need to check LHL, RHL and value of function at x = 1

Clearly,

f(1) = 1+1² = 2 [using equation 2]

LHL =

since 1–h is less than 1

∴ we will use the first expression from equation 2 i.e. f(x) = 1+x²

LHL =

RHL =

since 1+h is greater than 1

∴ we will use the second expression from equation 2 i.e. f(x) = 2–x

RHL =

Clearly, LHL ≠ RHL

∴ f(x) is discontinuous at x=1

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

2. Idea of sandwich theorem – This theorem also known as squeeze theorem that you may have encountered in your class 11 in limits chapter suggests that

If I be an interval having the point a as a limit point. Let g, f, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have{\displaystyle \lim _{x→ a}g(x)=\lim _{x→ a}h(x)=L.}{\displaystyle \lim _{x→ a}f(x)=L.}

g(x) ≤ f(x) ≤ h(x) and also

Then , We say that f(x) is squeezed between g(x) and h(x) or you can assume it like sandwich.

∴ …… equation 2

…… equation 3

NOTE : denominator in the above limits should be exactly same as that of content in sine function

Eg :

3. Idea of logarithmic limit and exponential limit –

……equation 4

…… equation 5

You must have read such limits in class 11. You can verify these by expanding log(1+x) and e^x in its taylor form.

Numerator and denominator conditions also hold for this limit like sandwich theorem.

E.g :

But,

Now we are ready to solve this problem:

Given function is :

…… equation 6

As we need to check continuity at x=0 , so we need to check value of f(x) at x = 0,left hand and right hand limits. If all 3 comes out to be equal we can say that f(x) is continuous at x=0 else it is discontinuous.

Clearly,

LHL =

Putting x = –h in equation 6, we have

LHL = [∵ sin(–θ) = sinθ and also tan(–θ) = tanθ ]

= [using equations 2 and 3 to apply sandwich theorem]

RHL =

Putting x = h in equation 6, we have

RHL = = [trying to apply logarithmic and exponential limit using equation 4 and 5]

= =

Clearly, LHL = RHL = f(0) =

∴ f(x) is continuous at x = 0.

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

2. Idea of sandwich theorem – This theorem also known as squeeze theorem that you may have encountered in your class 11 in limits chapter suggests that

If I be an interval having the point a as a limit point. Let g, f, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have{\displaystyle \lim _{x→ a}g(x)=\lim _{x→ a}h(x)=L.}{\displaystyle \lim _{x→ a}f(x)=L.}

g(x) ≤ f(x) ≤ h(x) and also

Then, We say that f(x) is squeezed between g(x) and h(x) or you can assume it like sandwich.

∴ …… equation 2

…… equation 3

NOTE : denominator in the above limits should be exactly same as that of content in sine function

Eg :

Let’s Solve :

…… equation 4

Given that f(x) is continuous at x=0

∴ LHL = RHL = f(0)

∴

From eqn 4:

f(0) =

LHL =

RHL =

=

= [∵ sinθ = 2sin(θ/2)cos(θ/2) and (1–cosθ) = 2sin²(θ/2)]

=

=

= [ using sandwich theorem as explained … eqn 2]

=

As LHL = RHL

∴ a =

∴ for f(x) to be continuous at x=0

a

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

Let’s Solve now:

Given function is

…… Equation 2

We need to check whether f(x) is continuous at x=0 or not

For this we need to check LHL, RHL and value of function at x=0

Clearly,

f(0) = 3*0 – 2 = –2 [from equation 2]

LHL =

RHL =

As, LHL ≠ RHL

∴ f(x) is discontinuous at x = 0

This can also be proved by plotting f(x) on cartesian plane.

For x >0 ,we need to plot

y = x + 1

put y=0, we get x=–1 and for second point we put x=0 and thus get y=1

two points are enough to plot the straight line.

Two coordinates are (–1,0) and (0,1)

For x≤0, we need to plot

y = 3x – 2

put x = 0 then y = –2

on putting y=0 we get x = 2/3

two coordinates are (0,–2) and ()

Graph:

It can be seen from graph that there is breakage in curve at (0,0)

Thus, it is discontinuous at x = 0

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

…… equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

Given,

…… equation 2

we are asked to check its continuity at x=0

∴ we need to check LHL, RHL and value of function at x = 0, if all comes out to be equal we can say f(x) is continuous at x=0 else it is discontinuous.

Clearly,

f(0) = 1 [from eqn 2]

LHL =

RHL =

Thus, LHL = RHL ≠ f(0)

∴ f(x) is discontinuous at x = 0

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

…… equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

Given,

…… equation 2

we are asked to check its continuity at x=1/2

∴ we need to check LHL ,RHL and value of function at x = 1/2 ,if all comes out to be equal we can say f(x) is continuous at x=1/2 else it is discontinuous.

Clearly,

f() = [from eqn 2]

LHL =

RHL =

Thus, LHL = RHL = f(0)

∴ f(x) is continuous at x =

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

…… equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

Given,

…… equation 2

we are asked to check its continuity at x=0

∴ we need to check LHL ,RHL and value of function at x = 0 ,if all comes out to be equal we can say f(x) is continuous at x=0 else it is discontinuous.

Clearly,

f(0) = 2*0+1 = 1 [from eqn 2]

LHL =

RHL =

Thus, LHL ≠ RHL

∴ f(x) is discontinuous at x =

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

…… equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

Given,

…… equation 2

We need to find the value of k such that f(x) is continuous at x = 1

Since f(x) is continuous at x = 1

∴ (LHL as x tends to 1) = (RHL as x tends to 1) = f(1)

∴

As, f(1) = k [from equation 2]

We can find either LHL or RHL to equate with f(1)

Let’s find RHL but if you want you can proceed with LHL also.

RHL =

=

As, f(x) is continuous

∴ RHL = f(1)

∴ k = 2

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

…… equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

Given,

…….equation 2

We need to find the value of k such that f(x) is continuous at x = 1

Since f(x) is continuous at x = 1

∴ (LHL as x tends to 1) = (RHL as x tends to 1) = f(1)

∴

As, f(1) = k [from equation 2]

We can find either LHL or RHL to equate with f(1)

Let’s find RHL,you can find LHL also.

RHL =

=

As, f(x) is continuous

∴ RHL = f(1)

∴ k = –1

Ideas required to solve the problem:

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

2. Idea of sandwich theorem – This theorem also known as squeeze theorem that you may have encountered in your class 11 in limits chapter suggests that

If I be an interval having the point a as a limit point. Let g, f, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have{\displaystyle \lim _{x→ a}g(x)=\lim _{x→ a}h(x)=L.}{\displaystyle \lim _{x→ a}f(x)=L.}

g(x) ≤ f(x) ≤ h(x) and also

Then , We say that f(x) is squeezed between g(x) and h(x) or you can assume it like sandwich.

∴ …… equation 2

NOTE : denominator in the above limit should be exactly same as that of content in sine function

Eg :

Given,

…….equation 3

We need to find the value of k such that f(x) is continuous at x = 0

Since f(x) is continuous at x = 0

∴ (LHL as x tends to 0) = (RHL as x tends to 0) = f(0)

∴

As, f(0) = k [from equation 2]

We can find either LHL or RHL to equate with f(1)

Let’s find RHL, you can proceed with LHL also.

RHL =

=

As, f(x) is continuous

∴ RHL = f(0)

∴ k =

Given:

It is clear that when x<2 and x>2, the given function is continuous at x = 2.

So, at x = 2

= 4k

We know that,

If f is continuous at x = c, then The Left–hand limit, the Right–hand limit and the value of the function at x = c exist and are equal to each other.

⇒ 4k = 3

k =

Therefore, the required value of k is

Given:

The function f is continuous at x = 0,Therefore,

We know that,

If f is continuous at x = c, then The Left–hand limit, the Right–hand limit and the value of the function at x = c exist and are equal to each other.

= 1

= 1

= k

Thus, f is continuous at x = 0 if k =

Given:

The function f is continuous at x = 2

We know that,

If f is continuous at x = c, then The Left–hand limit, the Right–hand limit and the value of the function at x = c exist and are equal to each other.

⇒ 2a + 5

since f is continuous at x = 2,

∴ 2a + 5 = 1

⇒ 2a = 1 – 5

⇒ 2a = –4

⇒ a = –2

Thus, f is continuous at x = 2 if a = –2.

To prove given f(x) is discontinuous at x = 0, we have to show that left–hand limit(LHL) and right–hand limit(RHL) is unequal.

LHL = , since (c–h)<c

RHL = = , since (c + h)>c

LHL

–1

RHL

1

since

The function f(x)remains discontinuous at x = 0, regardless the choice of k.

Given:

f(x) is continuous at x =

, since (c + h)>c

1

Also, given f() = 3

since f(x) is continuous at x =

i.e, = 3

k = 6

The value of k is 6 when f(x) is continuous at x =

Given:

f(x) is continuous at x = 0

For f(x) to be continuous at x = 0,f(0)^– = f(0) ⁺ = f(0)

LHL = f(0)^– =

1 (a + 1) + 1

(a + 1) + 1

f(0)^–a + 2 ...... (1)

RHL = f(0 ⁺ ) =

Take the complex conjugate of

,

i.e, and multiply it with numerator and denominator.

(a + b)(a–b) = a²– b²

f(0) ⁺ ...... (2)

since, f(x) is continuous at x = 0,From (1) & (2),we get,

a + 2 =

a = –2

a =

Also,

f(0)^–₌ f(0) ^{+ =} f(0)

f(0) = c

c = a + 2 =

c =

So the values of a = ,c = and b = R–{o}(any real number except 0 )

Given:

f(x) is continuous at x = 0 & f(0) =

For f(x) to be continuous at x = 0,f(0)^–₌ f(0) ^{+ =} f(0)

LHL = f(0)^– =

cos(–x) = cosx

f(0)^–

Since ,f(x) is continuous at x = 0 & f(0) =

k² = 1

k =

Given:

f(x) is continuous at x = 4 & f(4) = a + b

For f(x) to be continuous at x = 4, f(4)^–₌ f(4) ^{+ =} f(4)

LHL = f(4)^– =

a – 1

LHL = f(4) ⁺ =

1

Since ,f(x) is continuous at x = 4 & f(4) = a + b

f(4)^–=f(4) ⁺ =f(4)

a – 1 = a + b = 1

a – 1 = 1

a = 2

a + b = 1

2 + b = 1

b = 1–2

b = –1

For f(x) to be continuous at x = 0, f(0)^–₌ f(0) ^{+ =} f(0)

LHL = f(0)^– =

12

2

Since,

f(0)^–=f(0) ⁺ =f(0)

k = 2

For f(x) to be continuous at x = 0, f(0)^–= f(0) ⁺ =f(0)

LHL = f(0)^– =

hence,f(0) =

Given:

f(x) is continuous at x = 2 & f(2) = k

If f(x) to be continuous at x = 2,then,

f(2)^–₌ f(2) ^{+ =} f(2 )

LHL = f(2)^– =

(a + b)(a – b) = a²–b²

= 1

Since, f(x) is continuous at x = 2 and f(2) = k,then

k =

Given:

f(x) is continuous at x = 0 & f(0) = k

If f(x) to be continuous at x = 0,then

f(0)^–₌ f(0) ^{+ =} f(0)

LHL = f(0)^– =

[By Applying L – Hospital Rule.]

Hence, L.H.L = 0

As f(x) is continuous at x = 0.

Then, k = L.H.L

K = 0

Given:

f(x) is continuous at x = π,

If f(x) to be continuous at x = ,then

f(π)^–=f(π) ⁺ =f(π)

LHL

f() =

Given:

f(x) is continuous at x = 0

If f(x) to be continuous at x = 0,then

f(0)^–₌ f(0) ^{+ =} f(0)

LHL = f(0)^– =

= 1

⇒ 1

Hence,f(0) = 1

Given:

f(x) is continuous at x = 0 & f(0) = k

If f(x) to be continuous at x = 0,then,

f(0)^–₌ f(0) ^{+ =} f(0)

LHL = f(0)^– =

1

Since f(x) is continuous at x = 0 & f(0) = k,then

k = 1

Given:

f(x) is continuous at x = 0 & f(0) = 8

If f(x) to be continuous at x = 0,then,

f(0)^–₌ f(0) ^{+ =} f(0)

LHL = f(0)^– =

cos(–x) = cosx

cos2x = 1–2sin²x

1–cos2x = 2sin²x

1–cos2x = 2sin²x

2k²

Since f(x) is continuous at x = 0 & f(0) = 8,then

2 k² = 8

⇒ k² = 4

⇒ k = ±2

Given:

f(x) is continuous at x = 1 & f(1) = k

If f(x) to be continuous at x = 0,thenf(1)^–₌ f(1) ^{+ =} f(1)

LHL = f(1)^– =

tan(–x) = cotx

cos(0) = 1

Given:

f(x) is continuous at x = 0

If f(x) to be continuous at x = 0,thenf(0)^–₌ f(0) ^{+ =} f(0)

RHL = f(0)^– =

cos(0) = 1

f(0) ⁺

LHL = f(0)^– =

⇒ k×0

⇒ 0

Hence, the value of k = 0.

Given:

f(x) is continuous at x = 5

If f(x) to be continuous at x = 5,thenf(5)^–₌ f(5) ^{+ =} f(5)

LHL = f(5)^– =

⇒ 5k + 1

RHL = f(5) ⁺ =

⇒ 10

Since, f(x) is continuous at x = 5

we have,5k + 1 = 10

5k = 9

k =

The value of k is

Given:

f(x) is continuous at x = 5 & f(5) = k

If f(x) to be continuous at x = 5,thenf(5)^–= f(5) ⁺ =f(5)

LHL = f(5)^–

(a – b)² = a² – 2ab + b²

⇒ 10

Since ,f(x) is continuous at x = 5 & f(5) = k

k = 10

Given:

f(x) is continuous at x = 1

If f(x) to be continuous at x = 1,then,f(1)^–=f(1) ⁺ =f(1)

LHL = f(1)^– =

4 ....(1)

RHL = f(1) ⁺ =

⇒ k(1–0)²

⇒ k ....(2)

Since, f(x) is continuous at x = 1 & also

from (1) & (2)

k = 4

Given:

f(x) is continuous at x = 0

If f(x) to be continuous at x = 0,then,f(0)^–₌ f(0) ⁺₌ f(0)

LHL = f(0)^– =

k(0 + 2)

2k ...(1)

RHL = f(0) ⁺ =

1 ...(2)

Since, f(x) is continuous at x = 0,From (1) & (2),we get,

2k = 1

k =

Given:

f(x) is continuous at x = 2 & f(2) = k

If f(x) to be continuous at x = 2,then,f(2)^– = f(2) ⁺ = f(2) LHL = f(2)^– =

(a–b)³ = a³–b³–3a²b + 3ab²

(a–b)² = a²–2ab + b²

7

Since ,f(x) is continuous at x = 2 & f(2) = k

k = 7

Given:

f(x) is continuous at x = 3 & x = 5

If f(x) to be continuous at x = 3,then,f(3)^–₌ f(3) ⁺₌ f(3)

LHL = f(3)^– =

= 1 ...(1)

RHL = f(3) ⁺ =

⇒ a(3 + 0) + b

⇒ 3a + b ...(2)

Since ,f(x) is continuous at x = 3 and From (1) & (2),we get

3a + b = 1 ...(3)

Similarly ,f(x) is continuous at x = 5

If f(x) to be continuous at x = 5,then, f(5)^–₌ f(5) ⁺₌ f(5)

LHL = f(5)^– =

⇒ a(5–0) + b

⇒ 5a + b ...(4)

RHL = f(5) ⁺ =

7 ...(5)

Since , f(x) is continuous at x = 5 and From (4) & (5),we get,

5a + b = 7 ...(6)

Now equate (3) & (6)

⇒ a = 3

Now Substitute a = 3 in any one of above equation(3) & (6) ,

3a + b = 1

⇒ 3(3) + b = 1

⇒ 9 + b = 1

⇒ b = –8

Given:

For f(x) is continuous at x = 1

If f(x) to be continuous at x = 1,we have to show, f(1)^–=f(1) ⁺ = f(1)

LHL = f(1)^– =

(a–b)² = a²–2ab + b²

...(1)

RHL = f(1) ⁺ =

20² + 0 +

...(2)

From (1) & (2),we get f(1)^–₌ f(1) ⁺

Hence ,f(x) is continuous at x = 1

To prove whether f(x) is continuous at x = 0 & 1

If f(x) to be continuous at x = 0,we have to show that f(0)^–=f(0) ⁺ = f(0)

LHL = f(0)^– =

⇒ |(–0)| + |(–0)–1|

∴–|x| = |x| = x

⇒ |–1|

⇒ 1 ...(1)

RHL = f(0) ⁺ =

⇒ |0| + |0–1|

∴–|x| = |x| = x

⇒ |–1|

⇒ 1 ...(2)

From (1) & (2),we get f(0)^–=f(0) ⁺

Hence ,f(x) is continuous at x = 0

If f(x) to be continuous at x = 1,we have to show, f(1)^– =f(1) ⁺ = f(1)

LHL = f(1)^–₌

⇒ |(1–0) + (–0)|

⇒ |(1)|

⇒ 1 ...(3)

RHL = f(1) ^{+ =}

⇒ |(1 + 0)| + |0|

∴–|x| = |x| = x

⇒ |1|

⇒ 1 ...(4)

From (3) & (4),we get f(1)^–₌ f(1) ⁺

Hence ,f(x) is continuous at x = 1

To prove whether f(x) is continuous at –1 & 1

If f(x) to be continuous at x = –1,we have to show, f(–1)^–=f(–1) ⁺ =f(–1)

LHL = f(–1)^– =

⇒ |(–2–0)| + |–0|

⇒ |–2|

∴|–x| = |x| = x

⇒ 2 ...(1)

RHL = f(–1) ^{+ =}

⇒ |(–2 + 0)| + |0|

∴|–x| = |x| = x

|–2|

2 ...(2)

From (1) & (2),we get f(–1)^–=f(–1) ⁺

Hence ,f(x) is continuous at x = –1

If f(x) to be continuous at x = 1,we have to show, f(1)^– =f(1) ⁺ =f(1)

LHL = f(1)^– =

⇒ |–0| + |2–0|

⇒ |2|

⇒ 2 ...(3)

RHL = f(1) ⁺ =

⇒ |0| + |2 + 0|

∴|–x| = |x| = x

⇒ |2|

⇒ 2 ...(4)

From (3) & (4),we get f(1)^–=f(1) ⁺

Hence ,f(x) is continuous at x = 1

Given:

f(0) = 2

we have to prove that f(x) is discontinuous at x = 0

If f(x) to be discontinuous at x = 0,then

LHL = f(0)^– =

|–x| = |x| = x

⇒ 2 ...(1)

RHL = f(0) ⁺₌

⇒ 0 ...(2)

From (1) & (2),we know that,

f(0)^– f(0) ⁺

Hence, f(x) is discontinuous at x = 0

we have to find the value of 'k' such that f(x) is continuous at x = 0

If f(x) is be continuous at x = 0,then, f(0)^–=f(0) ⁺ = f(0)

LHL = f(0)^–=

⇒ –2×0² + k

⇒ k ...(1)

RHL = f(0) ⁺ =

⇒ 2×0² + k

⇒ k ...(2)

From (1) & (2),we get , f(0)^–₌ f(0) ⁺

So the value of k can be any real number(R),so that f(x) is continuous at x = 0

we have to find the value of '' such that f(x) is continuous at x = 0

If f(x) is be continuous at x = 0,then, f(0)^–=f(0) ⁺ =f(0)

LHL = f(0)^–=

⇒ (0² + 2×0)

⇒ 0 ...(1)

RHL = f(0) ⁺ =

⇒ 4(0) + 1

⇒ 1 ...(2)

From (1) & (2),we get f(0)^–=f(0) ⁺ ,

Hence f(x) is not continuous at x = 0

we also have to find out the continuity at point

For f(x) is be continuous at x = 1,

then ,f(0)^–=f(0) ⁺ =f(0)

LHL = f(1) ⁺ =

⇒ (0²–1)

⇒ – ...(1)

RHL = f(1) ^{+ =}

⇒ (5 + 4×0)

⇒ 5 ...(2)

From (1) & (2),we get f(0)^–₌ f(0) ⁺ ,

i.e, – = 5

⇒ = –5

Hence f(x) is continuous at x = 1,when = –5

Similarly, For f(x) is be continuous at x = –1,

then ,f(–1)^–=f(–1) ⁺ =f(–1)

LHL = f(–1)^–=

⇒ –(0² + 4×0 + 3)

⇒ –3 ...(3)

RHL = f(–1) ⁺ =

⇒ (–3 + 4×0)

⇒ –3 ...(2)

From (1) & (2),we get, f(–1)^–=f(–1) ⁺

i.e, –3 = –3

⇒ = 1

Hence f(x) is continuous at x = 1,when = 1

Given:

For f(x) is continuous at x = 2 & f(2) = k

If f(x) to be continuous at x = 2,we have to show, f(2)^–=f(2) ⁺ =f(2)

LHL = f(2)^–₌

⇒ (5–4×0)

⇒ 5 ...(1)

RHL = f(2) ⁺ =

⇒ (5 – 3 × 0)

⇒ 5 ...(2)

Since , f(x) is continuous at x = 2 & f(2) = k

k = 5

Given:

f(x) is continuous at x = & f() = a,

If f(x) to be continuous at x = ,we have to show, f()^–₌ f() ⁺₌ f()

LHL = f()^– =

sin(–x) = cosx

cos(–x) = sinx

(a³–b³) = (a–b)(a² + ab + b²)

∴ cos(0) = 1

...(1)

LHL = f() ⁺ =

1

...(2)

f(x) is continuous at x = & f() = a ,and from (1) & (2),we get

f()^–₌ f() ⁺₌ f()

= a

a =

⇒ b = 4

Hence ,a = & b = 4

we have to find the value of 'k'

Given:

f(x) is continuous at x = 0 & f(0) = k

If f(x) is be continuous at x = 0,then,

f(0)^–=f(0) ⁺ = f(0)

LHL = f(0)^– =

cos(0) = 1

⇒ (1)²

⇒ 1

RHL = f(0) ^{+ =}

⇒ 1

Since , f(x) is continuous at x = 0 & f(0) = k

And also , f(0)^–₌ f(0) ⁺₌ f(0)

So ,k = 1

we have to find the value of 'a' & 'b'

Given:

f(x) is continuous at x = 3

If f(x) is be continuous at x = 3,then,f(3)^–₌ f(3) ⁺₌ f(3)

LHL = f(3)^–₌

⇒ 3a – 0 × a + 1

⇒ 3a + 1 .....(1)

LHL = f(3) ⁺ =

⇒ 3b – 0 × b + 3

⇒ 3b + 3 ...(2)

Since ,f(x) is continuous at x = 3 and From (1) & (2),we get

3a + 1 = 3b + 3

⇒ 3a + 3b = 3 – 1

⇒ 3a + 3b = 2

⇒ 3(a + b) = 2

⇒ (a + b) =

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of the limit from class 11 we can summarise it as

A function is continuous at x = c if :

Here we have,

…….equation 1

To prove it everywhere continuous we need to show that at every point in the domain of f(x) [ domain is nothing but a set of real numbers for which function is defined ]

Clearly from definition of f(x) {see from equation 1}, f(x) is defined for all real numbers.

∴ we need to check continuity for all real numbers.

Let c is any random number such that c < 0 [thus c being a random number, it can include all negative numbers ]

f(c) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x < 0

Now, let m be any random number from the domain of f such that m > 0

thus m being a random number, it can include all positive numbers]

f(m) = m+1 [using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x > 0

As zero is a point at which function is changing its nature so we need to check LHL, RHL separately

f(0) = 0+1 = 1 [using eqn 1]

LHL =

[∵ sin – θ = – sin θ and ]

RHL =

Thus LHL = RHL = f(0).

∴ f(x) is continuous at x = 0

Hence, we proved that f is continuous for x < 0 ; x > 0 and x = 0

Thus f(x) is continuous everywhere.

Hence, proved.

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of the limit from class 11 we can summarise it as

A function is continuous at x = c if :

Here we have,

…….equation 1

The function is defined for all real numbers, so we need to comment about its continuity for all numbers in its domain ( domain = set of numbers for which f is defined )

Function is changing its nature (or expression) at x = 0, So we need to check its continuity at x = 0 first.

NOTE:

LHL = = =

[using eqn 1 and idea of mod fn]

RHL =

[using eqn 1 and idea of mod fn]

f(0) = 0

[using eqn 1]

Clearly, LHL ≠ RHL ≠ f(0)

∴ function is discontinuous at x = 0

Let c be any real number such that c > 0

∴ f(c) =

[using eqn 1]

And,

Thus,

∴ f(x) is continuous everywhere for x > 0.

Let c be any real number such that c < 0

∴ f(c) =

[using eqn 1 and idea of mod fn]

And,

Thus,

∴ f(x) is continuous everywhere for x < 0.

Hence, We can conclude by stating that f(x) is continuous for all Real numbers except zero.

Basic Concept:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…….equation 1

Function is defined for all real numbers so we need to comment about its continuity for all numbers in its domain ( domain = set of numbers for which f is defined )

Function is changing its nature (or expression) at x = 1, So we need to check its continuity at x = 1 first.

Clearly,

f(1) = 4 [using eqn 1]

Clearly,

∴ f(x) is discontinuous at x = 1.

Let c be any real number such that c ≠ 0

f(c) = c³ – c² + 2c – 2 [using eqn 1]

Clearly,

∴ f(x) is continuous for all real x except x =1

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…Equation 1

Note: [for changing the expression used identity:– (a²–b²) = (a+b)(a–b)]

Note: x – 2 is cancelled from numerator and denominator only because x ≠ 2, else we can’t cancel them

The function is defined for all real numbers, so we need to comment about its continuity for all numbers in its domain ( domain = set of numbers for which f is defined )

Function is changing its nature (or expression) at x = 2, So we need to check its continuity at x = 2 first.

Clearly,

f(2) = 16 [using eqn 1]

Note: (x – 2) is cancelled as x ≠ 2 but x → 2

Clearly,

∴ f(x) is continuous at x = 2.

Let c be any real number such that c ≠ 0

f(c) = [using eqn 1]

Clearly,

∴ f(x) is continuous for all real x

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…Equation 1

The function is defined for all real numbers, so we need to comment about its continuity for all numbers in its domain ( domain = set of numbers for which f is defined )

Let c is any random number such that c < 0 [thus c being a random number, it can include all negative numbers ]

f(c) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x < 0

Now, let m be any random number from the domain of f such that m > 0

thus m being a random number, it can include all positive numbers]

f(m) = 2m + 3 [using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x > 0

As zero is a point at which function is changing its nature so we need to check LHL, RHL separately

f(0) = 2×0+3 = 3 [using eqn 1]

LHL =

[∵ sin –θ = – sin θ and ]

RHL =

Thus LHL ≠ RHL

∴ f(x) is discontinuous at x = 0

Hence, f is continuous for all x ≠ 0 but discontinuous at x = 0.

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…Equation 1

The function is defined for all real numbers, so we need to comment about its continuity for all numbers in its domain ( domain = set of numbers for which f is defined )

Let c is any random number such that c ≠ 0 [thus c being a random number, it can include all numbers except 0 ]

f(c) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x ≠ 0

As zero is a point at which function is changing its nature, so we need to check the continuity here.

f(0) = 4 [using eqn 1]

LHL =

[∵ sin –θ = – sin θ and ]

RHL =

Thus LHL = RHL ≠ f(0)

∴ f(x) is discontinuous at x = 0

Hence, f is continuous for all x ≠ 0 but discontinuous at x = 0.

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…Equation 1

The function is defined for all real numbers, so we need to comment about its continuity for all numbers in its domain ( domain = set of numbers for which f is defined )

Let c is any random number such that c ≠ 0 [thus c being a random number, it can include all numbers except 0 ]

f(c) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x ≠ 0

As zero is a point at which function is changing its nature, so we need to check the continuity here.

f(0) = 5 [using eqn 1]

and,

[∵ ]

Thus

∴ f(x) is discontinuous at x = 0

Hence, f is continuous for all x ≠ 0 but discontinuous at x = 0.

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…Equation 1

The function is defined for all real numbers, so we need to comment about its continuity for all numbers in its domain ( domain = set of numbers for which f is defined )

Let c is any random number such that c ≠ 0 [thus c being a random number, it can include all numbers except 0 ]

f(c) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x ≠ 0

As zero is a point at which function is changing its nature so we need to check the continuity here.

f(0) = 10 [using eqn 1]

and,

[∵ using ]

Thus

∴ f(x) is discontinuous at x = 0

Hence, f is continuous for all x ≠ 0 but discontinuous at x = 0

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…Equation 1

Function is defined for all real numbers so we need to comment about its continuity for all numbers in its domain ( domain = set of numbers for which f is defined )

Let c is any random number such that c ≠ 0 [thus c being random number, it is able to include all numbers except 0 ]

f(c) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x ≠ 0

As x = 0 is a point at which function is changing its nature so we need to check the continuity here.

Since, f(0) = 7 [using eqn 1]

NOTE : Idea of logarithmic limit and exponential limit –

You must have read such limits in class 11. You can verify these by expanding log(1+x) and e^x in its taylor form.

Numerator and denominator conditions also hold for this limit like sandwich theorem.

E.g :

But,

and,

= [Using logarithmic and exponential limit as explained above, we have:]

=

Thus,

∴ f(x) is discontinuous at x = 0

Hence, f is continuous for all x ≠ 0 but discontinuous at x = 0

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

NOTE: Idea of modulus function |x| : You can think this function as a machine in which you can give it any real no. as an input and it returns its absolute value i.e. if positive is entered it returns the same no and if negative is entered it returns the corresponding positive no.

Eg:– |2| = 2 ; |–2| = –(–2) = 2

Similarly, we can define it for variable x, if x ≥ 0 |x| = x

If x < 0 |x| = (–x)

∴

Here we have,

Applying the idea of mod function, f(x) can be rewritten as:

……equation 1

Function is defined for all real numbers so we need to comment about its continuity for all numbers in its domain (domain = set of numbers for which f is defined )

Let c is any random number such that c < 1 [thus c being a random number, it can include all numbers less than 1]

f(c) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x < 1

As x = 1 is a point at which function is changing its nature, so we need to check the continuity here.

f(1) = | 1 – 3 | = 2 [using eqn 1]

LHL =

RHL =

Thus LHL = RHL = f(1)

∴ f(x) is continuous at x = 1

Now, again f(x) is changing its nature at x = 3, so we need to check continuity at x = 3

f(3) = 3– 3 = 0 [using eqn 1]

LHL =

RHL =

Thus LHL = RHL = f(3)

∴ f(x) is continuous at x = 3

For x > 3 ; f(x) = x–3 whose plot is linear, so it is continuous for all x > 3

You can verify it by checking limits.

Similarly, for 1 < x < 3, f(x) = 3–x whose plot is again a straight line and thus continuous for all point in this range.

Hence, f(x) is continuous for all real x.

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

NOTE: Idea of modulus function |x| : You can think this function as a machine in which you can give it any real no. as an input and it returns its absolute value i.e. if positive is entered it returns the same no and if negative is entered it returns the corresponding positive no.

Eg:– |2| = 2 ; |–2| = –(–2) = 2

Similarly, we can define it for variable x, if x ≥ 0 |x| = x

If x < 0 |x| = (–x)

∴

Here we have,

Applying the idea of mod function, f(x) can be rewritten as:

……equation 1

Function is defined for all real numbers so we need to comment about its continuity for all numbers in its domain (domain = set of numbers for which f is defined )

Let c is any random number such that c < –3 [thus c being random number, it is able to include all numbers less than –3]

f(c) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x < –3

As x = –3 is a point at which function is changing its nature so we need to check the continuity here.

f(–3) = 3–(–3) = 6 [using eqn 1]

LHL =

RHL =

Thus LHL = RHL = f(–3)

∴ f(x) is continuous at x = –3

Let c is any random number such that –3 < m < 3 [thus c being random number, it is able to include all numbers between –3 and 3]

f(c) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all –3 < x < 3

Now, again f(x) is changing its nature at x = 3,so we need to check continuity at x = 3

f(3) = 6*3+2 = 20 [using eqn 1]

LHL =

RHL =

Thus LHL ≠ RHL

∴ f(x) is discontinuous at x = 3

For x > 3 ; f(x) = 6x + 2 whose plot is linear, so it is continuous for all x > 3

You can verify it by checking limits.

Hence, f(x) is continuous for all real x except x = 3

There is only one point of discontinuity at x = 3

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

……equation 1

Function is defined for all real numbers so we need to comment about its continuity for all numbers in its domain (domain = set of numbers for which f is defined )

Let c is any random number such that c < 1 [thus c being random number, it is able to include all numbers less than 1]

f(c) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x < 1

As x = 1 is a point at which function is changing its nature so we need to check the continuity here.

f(1) = 1¹⁰ = 1 [using eqn 1]

LHL =

RHL =

Thus LHL = RHL = f(1)

∴ f(x) is continuous at x = 1

Let m is any random number such that m > 1 [thus m being random number, it is able to include all numbers greater than 1]

f(m) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all m > 1

Hence, f(x) is continuous for all real x

There no point of discontinuity. It is everywhere continuous

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

……equation 1

The function is defined for all real numbers, so we need to comment about its continuity for all numbers in its domain (domain = set of numbers for which f is defined )

Let c is any random number such that c < 0 [thus c being a random number, it can include all numbers less than 0]

f(c) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x < 0

As x = 0 is a point at which function is changing its nature, so we need to check the continuity here.

f(0) = 0

[using eqn 1]

LHL =

RHL =

Thus LHL = RHL = f(0)

∴ f(x) is continuous at x = 0

Let m is any random number such that 0 < m < 1 [thus m being a random number, it can include all numbers greater than 0 and less than 1]

f(m) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all 0 < x < 1

As x = 1 is again a point at which function is changing its nature, so we need to check the continuity here.

f(1) = 0

[using eqn 1]

LHL =

RHL =

Thus LHL ≠ RHL

∴ f(x) is discontinuous at x = 1

Let k is any random number such that k > 1 [thus k being a random number, it can include all numbers greater than 1]

f(k) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x > 1

Hence, f(x) is continuous for all real value of x, except x =1

There is a single point of discontinuity at x = 1

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…Equation 1

Function is defined for all real numbers so we need to comment about its continuity for all numbers in its domain ( domain = set of numbers for which f is defined )

Let c is any random number such that c ≠ 0 [thus c being a random number, it can include all numbers except 0 ]

f(c) = [ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x ≠ 0

As zero is a point at which function is changing its nature, so we need to check the continuity here.

f(0) = –1 [using eqn 1]

and,

Thus

∴ f(x) is continuous at x = 0

Hence, f is continuous for all x.

f(x) is continuous everywhere.

No point of discontinuity.

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

……….equation 1

Function is defined for all real numbers so we need to comment about its continuity for all numbers in its domain (domain = set of numbers for which f is defined)

For x < –1, f(x) is having a constant value, so the curve is going to be straight line parallel to x–axis.

So, it is everywhere continuous for x < –1.

It can be verified using limits as discussed in previous problems

Similarly for –1 < x < 1, plot on X–Y plane is a straight line passing through origin.

So, it is everywhere continuous for –1 < x < 1.

And similarly for x > 1, plot is going to be again a straight line parallel to x–axis

∴ it is also everywhere continuous for x > 1

From graph it is clear that function is continuous everywhere but let’s verify it with limits also.

As x = –1 is a point at which function is changing its nature so we need to check the continuity here.

f(–1) = –2 [using eqn 1]

LHL =

RHL =

Thus LHL = RHL = f(–1)

∴ f(x) is continuous at x = –1

Also at x = 1 function is changing its nature so we need to check the continuity here too.

f(1) = 2 [using eqn 1]

LHL =

RHL =

Thus LHL = RHL = f(1)

∴ f(x) is continuous at x = 1

Thus, f(x) is continuous everywhere and there is no point of discontinuity.

Basic Concept:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

Equation 1

Function is defined for all real numbers and we need to find the value of k so that it is continuous everywhere in its domain (domain = set of numbers for which f is defined)

As, for x ≠ 0 it is just a combination of trigonometric and linear polynomial both of which are continuous everywhere. It can be verified using limits and also by plotting curves. Since we are given that function is continuous everywhere so don’t need to bother about that.

As x = 0 is only point at which function is changing its nature so it needs to be continuous here.

f(0) = 3k [using eqn 1]

and,

[∵ ]

∵ f(x) is continuous everywhere [given in question]

∴ 3k =

∴ k =

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…………………..equation 1

Function is defined for all real numbers and we need to find the value of k so that it is continuous everywhere in its domain (domain = set of numbers for which f is defined)

To find the value of constants always try to check continuity at the values of x for which f(x) is changing its expression.

As most of the time discontinuities are here only, if we make the function continuous here, it will automatically become continuous everywhere

From equation 1 ,it is clear that f(x) is changing its expression at x = 2

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering RHL as RHL will give expression independent of k]

[using equation 1]

∴ 2k + 5 = 1

2k = –4

k = = –2

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…………………..equation 1

Function is defined for all real numbers and we need to find the value of k so that it is continuous everywhere in its domain (domain = set of numbers for which f is defined)

To find the value of constants always try to check continuity at the values of x for which f(x) is changing its expression.

As most of the time discontinuities are here only, if we make the function continuous here, it will automatically become continuous everywhere

From equation 1 ,it is clear that f(x) is changing its expression at x = 0

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering LHL as LHL will give expression dependent of k]

[using equation 1]

∵ k * 0 = 1

As above equality never holds true for any value of k

k = not defined

No such value of k is possible for which f(x) is continuous everywhere.

F(x) will always have a discontinuity at x = 0

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…………………..equation 1

Function is defined for all real numbers and we need to find the value of a & b so that it is continuous everywhere in its domain (domain = set of numbers for which f is defined)

To find the value of constants always try to check continuity at the values of x for which f(x) is changing its expression.

As most of the time discontinuities are here only, if we make the function continuous here, it will automatically become continuous everywhere

From equation 1 ,it is clear that f(x) is changing its expression at x = 3

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering RHL as RHL will give expression inclusive of a & b]

[using equation 1]

∴ 3a + b = 2 ……………….Equation 2

Also from equation 1 ,it is clear that f(x) is also changing its expression at x = 5

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering LHL as LHL will give expression inclusive of a & b]

[using equation 1]

∴ 5a + b = 9 ……………….Equation 3

As , b = 9 – 5a

Putting value of b in equation 2:

3a + 9 – 5a = 2

2a = 7

a =

∴ b = 9 – 5(

∴ a = and b =

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

we have,

…………………..equation 1

Function is defined for all real numbers and we need to find the value of a & b so that it is continuous everywhere in its domain (domain = set of numbers for which f is defined)

To find the value of constants always try to check continuity at the values of x for which f(x) is changing its expression.

As most of the time discontinuities are here only, if we make the function continuous here, it will automatically become continuous everywhere

From equation 1 ,it is clear that f(x) is changing its expression at x = –1

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering RHL as RHL will give expression inclusive of a & b]

[using equation 1]

∴ a + b = 4 ……………….Equation 2

Also from equation 1, it is clear that f(x) is also changing its expression at x = 0

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering LHL as LHL will give expression inclusive of a & b]

[using equation 1]

∴ b = 1 ……………….Equation 3

Putting value of b in equation 2:

a + 1 = 4

a = 3

∴ a = and b =

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

……………..equation 1

Function is defined for all real numbers and we need to find the value of p so that it is continuous everywhere in its domain (domain = set of numbers for which f is defined)

To find the value of constants always try to check continuity at the values of x for which f(x) is changing its expression.

As most of the time discontinuities are here only, if we make the function continuous here, it will automatically become continuous everywhere

From equation 1 ,it is clear that f(x) is changing its expression at x = 0

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering LHL as LHL will give expression inclusive of p]

[using equation 1]

∴ p =

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…………………..equation 1

Function is defined for all real numbers and we need to find the value of a & b so that it is continuous everywhere in its domain (domain = set of numbers for which f is defined)

To find the value of constants always try to check continuity at the values of x for which f(x) is changing its expression.

As most of the time discontinuities are here only, if we make the function continuous here, it will automatically become continuous everywhere

From equation 1 ,it is clear that f(x) is changing its expression at x = 2

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering RHL as RHL will give expression inclusive of a & b]

[using equation 1]

∴ 2a + b = 5 ……………….Equation 2

Also from equation 1 ,it is clear that f(x) is also changing its expression at x = 10

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering LHL as LHL will give expression inclusive of a & b]

[using equation 1]

∴ 10a + b = 21 ……………….Equation 3

As , b = 21 – 10a

Putting value of b in equation 2:

2a + 21 – 10a = 5

8a = 16

a =

∴ b = 21 – 10×2

∴ a = 2 and b =

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

……………..equation 1

Function is defined for all real numbers and we need to find the value of k so that it is continuous everywhere in its domain (domain = set of numbers for which f is defined)

To find the value of constants always try to check continuity at the values of x for which f(x) is changing its expression.

As most of the time discontinuities are here only, if we make the function continuous here, it will automatically become continuous everywhere

From equation 1 ,it is clear that f(x) is changing its expression at x = π/2

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering LHL as LHL will give expression inclusive of k]

[using equation 1]

[∵

∴ k = 3×2 = 6

∴ k =

Basic Concept:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…………………..equation 1

The function is defined for [0,∞] and we need to find the value of a and b so that it is continuous everywhere in its domain (domain = set of numbers for which f is defined)

To find the value of constants always try to check continuity at the values of x for which f(x) is changing its expression.

As most of the time discontinuities are here only, if we make the function continuous here, it will automatically become continuous everywhere

From equation 1 ,it is clear that f(x) is changing its expression at x = 1

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering LHL as LHL will give expression inclusive of a ]

[using equation 1]

∴

∴ a = ± 1 …………… equation 2

Also from equation 1 ,it is clear that f(x) is also changing its expression at x = √2

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering LHL as LHL will give expression inclusive of a & b]

[using equation 1]

∴ b² – 2b = a ……………….Equation 3

From equation 2, a = –1

b² – 2b = –1

⇒ b² – 2b + 1 = 0

⇒ (b – 1)² = 0

∴ b = 1 when a = –1

Putting a = 1 in equation 3:

b² – 2b = 1

⇒ b² – 2b – 1 = 0

⇒

Thus,

For a = –1 ; b = 1

For a = 1 ; b = 1 ± √2

Basic Concept:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…………………..equation 1

Function is defined for [0,π] and we need to find the value of a and b so that it is continuous everywhere in its domain (domain = set of numbers for which f is defined)

To find the value of constants always try to check continuity at the values of x for which f(x) is changing its expression.

As most of the time discontinuities are here only, if we make the function continuous here, it will automatically become continuous everywhere

From equation 1, it is clear that f(x) is changing its expression at x = π/4

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering LHL as LHL will give expression inclusive of a & b]

[using equation 1]

∴

∴ a + b = – π/4 …………… equation 2

Also from equation 1 ,it is clear that f(x) is also changing its expression at x = π/2

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering LHL as LHL will give expression inclusive of a & b]

[using equation 1]

b = – a – b

∴ a = –2b ……………….Equation 3

Putting value of a from equation 3 to equation 2

∴ –2b + b = – π/4

⇒ b = π/4

∴ a = –2×(π/4)

= –π/2

Thus, a = –π/2 and b = π/4

Basic Concept:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…………………..equation 1

Function is defined for [0,8] and we need to find the value of a and b so that it is continuous everywhere in its domain (domain = set of numbers for which f is defined)

To find the value of constants always try to check continuity at the values of x for which f(x) is changing its expression.

As most of the time discontinuities are here only, if we make the function continuous here, it will automatically become continuous everywhere

From equation 1, it is clear that f(x) is changing its expression at x = 2

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering LHL as LHL will give expression inclusive of a & b]

[using equation 1]

4+2a + b = 8

∴ 2a + b = 4

∴ b = 4 – 2a …………… equation 2

Also from equation 1 ,it is clear that f(x) is also changing its expression at x = 4

Given,

f(x) is continuous everywhere

[using basic ideas of limits and continuity]

[considering RHL as RHL will give expression inclusive of a & b]

[using equation 1]

∴ 8a + 5b = 14 ……………….Equation 3

Putting value of a from equation 2 to equation 3

∴ 8a + 5(4–2a) = 14

⇒ 2a = 6

∴ a = 6/2

= 3

∴ b = 4 – 2×3 = –2

Thus, a = 3 and b = –2

Basic Concept:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Function is defined for [0,π] and we need to find the value of f(x) so that it is continuous everywhere in its domain (domain = set of numbers for which f is defined)

As we have expression for x ≠ π/4, which is continuous everywhere in [0,π],so

If we make it continuous at x = π/4 it is continuous everywhere in its domain.

Given,

……….equation 1

Let f(x) is continuous for x = π/4

∴

= = [∵ tan (π/2–θ) = cot θ ]

= [multiplying and dividing by π/4–x and π/2–2x to apply sandwich theorem]

=

[∵

∴ value that can be assigned to f(x) at x = π/4 is

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…….equation 1

Function is defined for all real numbers so we need to comment about its continuity for all numbers in its domain ( domain = set of numbers for which f is defined )

Function is changing its nature (or expression) at x = 2, So we need to check its continuity at x = 2 first.

LHL = = = [using eqn 1]

RHL = [using eqn 1]

f(2) =

[using eqn 1]

Clearly, LHL = RHL = f(2)

∴ function is continuous at x = 2

Let c be any real number such that c > 2

∴ f(c) = [using eqn 1]

And,

Thus,

∴ f(x) is continuous everywhere for x > 2.

Let m be any real number such that m < 2

∴ f(m) = [using eqn 1]

And,

Thus,

∴ f(x) is continuous everywhere for x < 2.

Hence, We can conclude by stating that f(x) is continuous for all Real numbers

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of the limit from class 11 we can summarise it as a function is continuous at x = c if :

NOTE:

Here we are given with

f(x) = sin |x|

…………………………equation 1

Function is defined for all real numbers so we need to comment about its continuity for all numbers in its domain ( domain = set of numbers for which f is defined )

Function is changing its nature (or expression) at x = 0, So we need to check its continuity at x = 0 first.

LHL = =

[using eqn 1]

RHL = [using eqn 1]

f(0) =

[using eqn 1]

Clearly, LHL = RHL = f(0)

∴ function is continuous at x = 0

For all x ≠ 0, f(x) is simply a trigonometric function which is everywhere continuous which can be verified by seeing its plot in X–Y plane or even can be verified using limit.

∴ f(x) is everywhere continuous in its domain i.e. it is continuous for all real values of x

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…….equation 1

To prove it everywhere continuous we need to show that at every point in domain of f(x) [ domain is nothing but a set of real numbers for which function is defined ]

Clearly from definition of f(x) { see from equation 1}, f(x) is defined for all real numbers.

∴ we need to check continuity for all real numbers.

Let c is any random number such that c < 0 [thus c being a random number, it can include all negative numbers ]

f(c) =

[ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x < 0

Now, let m be any random number from domain of f such that m > 0

thus m being a random number, it can include all positive numbers]

f(m) = m+1 [using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x > 0

As zero is a point at which function is changing its nature so we need to check LHL, RHL separately

f(0) = 0+1 = 1 [using eqn 1]

LHL = [∵ sin –θ = sin θ and ]

RHL =

Thus LHL = RHL = f(0).

∴ f(x) is continuous at x = 0

Hence, we proved that f is continuous for x < 0 ; x > 0 and x = 0

Thus f(x) is continuous everywhere.

Hence, proved.

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Given,

g(x) = x – [x] …………equation 1

We need to prove that g(x) is discontinuous at every integral point.

Note: Idea of greatest integer function [x] –

Greatest integer function can be seen as an input output machine in which if you enter a number, It returns the greatest integer that is just less than number x.

For example : [2.5] = 2 ; [9.99998] = 9 ; [–3.899] = –4 ; [4] = 4

To prove g(x) discontinuous at integral points we need to show that

Where c is any integer.

Let c is any integer

∴ g(c) = c – [c] = c – c =0 [using eqn 1]

LHL =

[∵ c is integer and h is very small positive no, so (c–h) is a number less than integer c ∴ greatest integer less than (c –h) = (c–1)]

RHL = [using eqn 1 and idea of gif function]

Thus, LHL ≠ RHL

∴ g(x) is discontinuous at every integral point.

Idea : If f and g are two functions whose domains are same and both f and g are everywhere continuous then :

i) f + g is also everywhere continuous

ii) f – g is also everywhere continuous

iii) f*g is also everywhere continuous

∵ f(x) = sin x + cos x

It is a purely trigonometric function

As sin x is continuous everywhere and cos x is also continuous everywhere for all real values of x

As f(x) is nothing but sum of two everywhere continuous function

∴ f(x) is also everywhere continuous.

We can see this through its graph which shows no point of discontinuity.

Fig : plot of sin x + cos x

Idea : If f and g are two functions whose domains are same and both f and g are everywhere continuous then :

i) f + g is also everywhere continuous

ii) f – g is also everywhere continuous

iii) f*g is also everywhere continuous

∵ f(x) = sin x – cos x

It is a purely trigonometric function

As sin x is continuous everywhere and cos x is also continuous everywhere for all real values of x

As f(x) is nothing but difference of two everywhere continuous function

∴ f(x) is also everywhere continuous.

We can see this through its graph which shows no point of discontinuity.

Fig : plot of sin x + cos x

Idea : If f and g are two functions whose domains are same and both f and g are everywhere continuous then :

i) f + g is also everywhere continuous

ii) f – g is also everywhere continuous

iii) f*g is also everywhere continuous

iv) f/g is also everywhere continuous for all R except point at which g(x) = 0

∵ f(x) = sin x × cos x

It is a purely trigonometric function

As sin x is continuous everywhere and cos x is also continuous everywhere for all real values of x

As f(x) is nothing but product of two everywhere continuous function

∴ f(x) is also everywhere continuous.

We can see this through its graph which shows no point of discontinuity.

Fig : plot of sin x × cos x

Idea: Such problems can be solved easily using the idea of the continuity of composite function.

If we not go to very strict mathematical meaning of composite function you can think it as it is a function of function.

Let, g(x) = cos x

And h(x) = x²

Then g(h(x)) = g(x²) = cos x²

We write g(h(x)) as (goh)(x) and this is what we called composite function/function composition.

We have a theorem regarding composition of function in continuity which lets us to solve problems easily.

Theorem: If f and g are real valued function such that (fog) is defined at c, and g is continuous at c and f is continuous at g(c) then (fog) is continuous at x = c

For our problem:

Let, g(x) = cos x

and h(x) = x²

Given: f(x) = cos x² = g(h(x)) = (goh)(x)

Clearly, h(x) is a polynomial function, which is everywhere continuous

And g(x) being cosine function, it is also everywhere continuous.

FIG : Plot of cos x²

∴ goh(x) = f(x) is also everywhere continuous. [using above explained theorem]

Idea : Such problems can be solved easily using idea of the continuity of composite function.

If we not go to very strict mathematical meaning of composite function you can think it as it is a function of function.

Let, g(x) = cos x

And h(x) = x²

Then g(h(x)) = g(x²) = cos x²

We write g(h(x)) as (goh)(x) and this is what we called composite function/function composition.

We have a theorem regarding composition of function in continuity which lets us to solve problems easily.

Theorem: If f and g are real valued function such that (fog) is defined at c, and g is continuous at c and f is continuous at g(c) then (fog) is continuous at x = c

For our problem:

Let, g(x) = |x|

and h(x) = cos x

Given: f(x) = |cos x| = g(h(x)) = (goh)(x)

Clearly, h(x) is a cosine(trigonometric) function, which is everywhere continuous

And g(x) being mod function ,it is also everywhere continuous.

∴ goh(x) = f(x) is also everywhere continuous. [using above explained theorem]

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of the limit from class 11 we can summarise it as a function is continuous at x = c if :

NOTE:

Here we have,

f(x) = |x| – |x + 1|

f(x) rewritten using idea of mod function:

…….equation 1

Clearly for x < –1 , f(x) = constant and also for x > 1 f(x) is constant

∴ in these regions f(x) is everywhere continuous.

For –1 < x < 0, plot of graph is a straight line as in this region f(x) is given by linear polynomial

∴ it is also continuous here.

∵ function is changing its expression at x = –1 and x = 0, so we need to check continuities at these points.

At x = –1 :

f(–1) = 1 [using equation 1]

LHL = [using equation 1]

RHL =

[using equation 1]

Clearly,

LHL = RHL = f(–1)

∴ it is continuous at x = –1

At x = 0 :

f(0) = –2*0–1 = –1 [using equation 1]

LHL =

[using equation 1]

RHL =

[using equation 1]

Clearly,

LHL = RHL = f(0)

∴ it is continuous at x = 0

Hence,

f(x) is continuous everywhere in its domain.

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of the limit from class 11 we can summarise it as a function is continuous at x = c if :

Given :

……………….Equation 1

As for x ≠ 0, f(x) is just a product of two everywhere continuous function

∴ it is continuous for all x ≠ 0.

∵ f(x) is changing its nature at x = 0, So we need to check continuity at x = 0

f(0) = 0 [using equation 1]

and = 0

[∵ sin(1/0) is also going to be a value between [–1,1] ,so its product with 0 = 0]

Thus,

∴ It is continuous at x = 0

Hence, it is everywhere continuous.

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

NOTE: If f and g are two functions whose domains are same and both f and g are everywhere continuous then f/g is also everywhere continuous for all R except point at which g(x) = 0

As, f(x) =

Domain of f = { all Real numbers except 2 } = R – {–2}

Clearly it is not defined at x = –2, for rest of values it is continuous everywhere

Because 1 is everywhere continuous and x + 2 is also everywhere continuous

∴ f(x) is everywhere continuous except at x = –2

f(f(x)) = f

Domain of f(f(x)) = R – {–2 , (}

For rest of values it just a fraction of two everywhere continuous function

∴ at all other points it is everywhere continuous.

Hence,

f(f(x)) is discontinuous at x = –2 and x = –5/2

Clearly,

is discontinuous at x = 1 as t is not defined at this point.

∴ f(t) being a composition of function involving t, it is also discontinuous at x = 1

∵

by observing f(t) we can say that f(t) is not defined at

t = – 2 and t = 1.

∴ this will also contribute to discontinuity

Hence,

⇒ x – 1 = –1/2

⇒ x = 1 – 1/2

⇒ x = 1/2

also other point of discontinuity is obtained by :

⇒ x – 1 = 1

⇒ x = 2

Hence points at which f(t) is discontinuous are x = { 1/2,1,2}

At all other point it is continuous.

Formula:- (i) A function f(x) is said to be continuous at a point x=a of its domain, iff

f(x)is discontinuous for three point

Formula:-

(i) A function f(x) is said to be continuous at a point x=a of its domain, iff

(ii) for right hand derivative

) for left hand derivative

Given:-

f(x) = |x – a| ϕ(x)

using formula (ii)

For L.H.L

Formula:-

(i) A function f(x) is said to be continuous at a point x=a of its domain, iff

(ii) for right hand derivative

) for left hand derivative

Given:-

f(x) = |log₁₀ x|

Using limit at x=1