Limits

Given

f(x) =

f(x) =

To find

To limit to exist, we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

From above equations

(from 2)

Thus, limit does not exist.

Given f(x) =

To find

To limit to exist, we know …….(1)

thus

From (1)

2(2 + 0) + 3 = (2 - 0) + k

4 + 3 = 2 + k

5 = k

To find

To limit to exist, we know …….(1)

Thus, to find the limit using the concept ……(2)

……..(3)

……(4)

From above equations

Thus, limit does not exist.

Given f(x) =

f(x) =

f(x) =

To find

To limit to exist we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

From above equations

Thus, limit does not exist.

Given f(x) =

To find whether exists?

To limit to exist we know …….(1)

Thus to limit to exist ……(2)

From above equations

Thus, the limit does not exists.

Given f(x) =

To find whether exists?

To limit to exist we know …….(1)

Thus to limit to exist ……(2)

From above equations

Thus, the limit does not exists.

Given f(x) =

To find

To limit to exist we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

From above equations

Thus from (2),(3) and (4)

Given f(x) =

(i)To find

To limit to exist, we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

…….(5)

From above equations

thus the limit exists

Thus from (5)

(ii) To find

To limit to exist, we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

From above equations

Thus from (2),(3) and (4)

Given f(x) =

To find

To limit to exist we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

From above equations

thus the limit does not exists

Given f(x) =

f(x) =

f(x) =

To find

To limit to exist we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

From above equations

Thus limit does not exists

Given: f(x) = (x – a₁)(x – a₂)…..(x – a_n)

Now,

Given f(x) =

To find

Given f(x) =

To find

Given f(x) =

To find

Given f(x) =

To find

Given f(x) =

Factorizing f(x)

f(x) =

f(x) =

f(x) =

To find

Given f(x) =

To find

Some standard limit are:

Thus to find:

= ∞

Some standard limit are:

Thus to find:

= - ∞

Given f(x) =

Factorizing f(x)

f(x) =

f(x) =

f(x) =

f(x) =

To find

Some standard limit are:

Thus to find:

= 2 + ∞ = ∞

Some standard limit are:

Thus to find:

= 1 - ∞ = - ∞

Given f(x) =

To find

To limit to exist we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

From above equations

Thus, limit does not exist.

We know greatest integer [x] is the integer part.

For f(x) = [x]

To find:

To limit to exist we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

From above equations

Thus, the limit does not exist.

We know greatest integer [x] is the integer part.

For f(x) = [x]

To find:

To limit to exist we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

From above equations

Thus, limit does exists.

We know greatest integer [x] is the integer part.

For f(x) = [x]

To find:

To limit to exist we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

From above equations

Thus limit does not exists.

To Prove:

L.H.S (Since, [a + h] = [a])

Hence, Proved.

Also,

To prove:

L.H.S (Since, [1 – h] = 0)

Hence, Proved.

We know greatest integer [x] is the integer part.

For f(x) = x/[x]

To show

Proof:

To limit to exist we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

From above equations

We know greatest integer [x] is the integer part.

For f(x) = x/[x]

To show

Proof:

To limit to exist we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

From above equations

We know greatest integer [x] is the smallest integer nearest to that number .

For f(x) = [x]

To find:

To limit to exist we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

From above equations

Thus limit does exists

Given f(x) =

To find

To limit to exist we know …….(1)

Thus to find the limit using the concept ……(2)

……..(3)

……(4)

…..(5)

From above equations

Thus the limit does not exist

To Prove: does not exist

Let us take the left-hand limit for the function:

L.H.L

Now, multiplying and dividing by h, we get,

L.H.L

Now, taking the right-hand limit of the function, we get,

R.H.L

Now, multiplying and dividing by h, we get,

R.H.L

Clearly, L.H.L ≠ R.H.L

Hence, limit does not exist.

Let us find the limit of the function at .

Let

Therefore,

L.H.L

Now,

Hence, k = 6.

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z = (indeterminate form)

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

Note: While modifying be careful that you don’t introduce any zero terms in the denominator

As Z

Multiplying both numerator and denominator by √(4+x)+2 so that we can remove the indeterminate form.

∴ Z

⇒ Z

{using a² - b² = (a + b)(a - b)}

⇒ Z

Using basic algebra of limits-

Z =

⇒ Z = 4

Use the formula:

∴ Z = 4log 5

Or,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential and logarithmic limits.

Use the formula: and

As Z =

To get the above forms, we need to divide numerator and denominator by x.

∴ Z {using basic limit algebra}

⇒ Z {using the formulae described above}

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

As Z =

∴ Z = {using (a+b)² = a²+b²+2ab}

Using algebra of limit, we can write that

Z =

Use the formula:

∴ Z =

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential and logarithmic limits.

To get the desired forms, we need to include mx and nx as follows:

∴ Z

⇒ Z

Using algebra of limits-

Use the formula:

∴ Z =

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential and logarithmic limits.

To get similar forms as in a formula, we move as follows-

As Z

⇒ Z

Using algebra of limits we have-

Z

Use the formula:

∴

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

As Z

∴ Z

{using (a-b)² = a²+b²-2ab}

Z

To apply the formula we need to bring the exact form present in the formula, so-

Z

{Adding and subtracting 1 in numerator}

⇒ Z

Using algebra of limits-

Z

Use the formula:

∴ Z =

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

As Z

Using Algebra of limits-

We have-

Z

Use the formula:

∴ Z = log 4 × log 2

∵ log 4 = log 2² = 2log 2

{using properties of log}

∴ Z = 2(log 2)²

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential and logarithmic limits.

To get the desired forms, we need to include mx and nx as follows:

As Z

⇒ Z

{Adding and subtracting 1 in numerator}

⇒ Z

{using algebra of limits}

To get the form as present in the formula we multiply and divide m and n into both terms respectively:

∴ Z

Use the formula:

∴ Z = m log a – n log b

{using properties of log}

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential and logarithmic limits.

To get similar forms as in a formula, we move as follows-

As Z

⇒ Z

Using algebra of limits we have-

Z

Use the formula:

∴

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential and logarithmic limits.

To get similar forms as in a formula, we move as follows-

As x→2 ∴ x-2 →0

Let x-2 = y

∴ Z =

We can’t use the formula directly as the base of log is we need to change this to e.

Applying the formula for change of base-

We have-

∴ Z =

Use the formula:

∴ Z =

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential and logarithmic limits.

To get similar forms as in formula, we move as follows-

As Z

⇒ Z

Using algebra of limits we have-

Z

Use the formula:

∴

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential and logarithmic limits.

To get similar forms as in formula, we move as follows-

Let 1/x = y

As x→∞ ⇒ y→ 0

∴ Z can be rewritten as-

Z

Use the formula:

∴ Z =

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential and logarithmic limits and also use of sandwich theorem -

To get the desired forms, we need to include mx and nx as follows:

As Z

⇒ Z {Adding and subtracting 1 in numerator}

⇒ Z

{using algebra of limits}

To get the form as present in the formula we multiply and divide x into both terms respectively:

∴ Z

{manipulating to get the forms present in formulae}

Z

Use the formula: and

∴ Z

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential and logarithmic limits.

To get similar forms as in a formula, we move as follows-

As Z =

⇒ Z =

Using algebra of limits we have-

Z =

Use the formula:

∴

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-

As Z

⇒ Z

Use the formula: and

∴ Z = log e + 1

{∵ log e = 1}

⇒ Z = 1+1 = 2

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-

As Z

To get the desired form to apply the formula we need to divide numerator and denominator by x.

⇒ Z

Using algebra of limits, we have-

Z

Use the formula: and

∴ Z =

{∵ log e = 1}

⇒ Z = 2

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-

As Z

To get rid of indeterminate form we will divide numerator and denominator by sin x

∴ Z

Using Algebra of limits we have-

Z

Where, A

and B

{from sandwich theorem}

As A

Let, sin x =y

As x→0 ⇒ y→0

∴ A

Using

A = log e = 1

∴

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-

As Z =

Adding and subtracting 1 in the numerator to get the desired form

⇒ Z =

⇒ Z =

{using algebra of limits}

To get the desired form to apply the formula we need to divide numerator and denominator by x.

⇒ Z

Using algebra of limits, we have-

Z

Use the formula: and

∴ Z =

{∵ log e = 1}

⇒ Z = 1/2

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

As Z

To apply the formula of logarithmic limits we need to get the form that matches with one in formula

∴ We proceed as follows-

Z

⇒ Z

⇒ Z

∵ x→a ⇒ x/a →1

⇒ x/a – 1 → 0

Let, (x/a)-1 = y

∴ y→0

Hence, Z can be rewritten as-

Z

Use the formula:

∴ Z

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

As Z =

To apply the formula of logarithmic limits we need to get the form that matches with one in formula

∴ We proceed as follows-

Z =

⇒ Z =

⇒ Z =

To apply the formula of logarithmic limit we need denominator

∴ multiplying in numerator and denominator

Hence, Z can be rewritten as-

Z =

⇒ Z =

{Using algebra of limits}

⇒ Z =

As, x→0 ⇒

Let,

∴ Z =

Use the formula:

∴ Z =

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

As Z =

To apply the formula of logarithmic limits we need to get the form that matches with one in formula

∴ We proceed as follows-

Z =

{using properties of log}

⇒ Z =

To apply the formula of logarithmic limit, we need the x/2 denominator

∴ multiplying 1/2 in numerator and denominator

Hence, Z can be rewritten as-

Z =

⇒ Z =

{Using algebra of limits}

As x→0 ⇒

Let,

∴ Z =

Use the formula:

∴ Z =

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

As Z =

To apply the formula of logarithmic limits we need to get the form that matches with one in formula

∴ We proceed as follows-

Z = {using properties of log}

⇒ Z =

To apply the formula of logarithmic limit, we need x/a in the denominator

∴ multiplying 1/a in numerator and denominator

Hence, Z can be rewritten as-

Z =

⇒ Z =

{Using algebra of limits}

As x→0 ⇒

Let,

∴ Z =

Use the formula:

∴ Z =

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

As Z =

To apply the formula of logarithmic limits we need to get the form that matches with one in formula

∴ We proceed as follows-

Z =

⇒ Z =

⇒ Z =

To apply the formula of logarithmic limit we need denominator

∴ multiplying in numerator and denominator

Hence, Z can be rewritten as-

Z =

⇒ Z =

{Using algebra of limits}

⇒ Z =

As, x→0 ⇒

Let,

∴ Z =

Use the formula:

∴ Z =

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential and logarithmic limits.

As Z =

⇒ Z =

{Adding and subtracting 1 in numerator}

⇒ Z =

{using algebra of limits}

Use the formula:

∴ Z = log 8 – log 2 =

{using properties of log}

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-

As Z =

As, 1-cos x = 2sin²(x/2)

∴ Z =

⇒ Z =

To get the desired form to apply the formula we need to divide numerator and denominator by x².

⇒ Z =

Using algebra of limits, we have-

Z =

Use the formula: and

∴ Z =

⇒ Z = 2 log 2

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

As Z =

To apply the formula of logarithmic limits we need to get the form that matches with one in formula

∴ multiplying numerator and denominator by

⇒ Z =

⇒ Z =

{using (a+b)(a-b)=a²-b²}

⇒ Z =

⇒ Z =

Use the formula:

∴ Z = 1/2

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-

As Z =

To apply the formula of logarithmic limits we need to get the form that matches with one in formula

∴ dividing numerator and denominator by x³

⇒ Z =

⇒ Z =

⇒ Z =

{using algebra of limits}

Use the formula: and

∴ Z = 1/1

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential limits.

As Z =

⇒ Z =

⇒ Z =

{using properties of exponents}

⇒ Z =

{using algebra of limits}

⇒ Z =

∴ Z =

As, x→ (π/2)

∴ cot(π/2) – cos(π/2) → 0

Let, y = cot x – cos x

∴ if x→π/2 ⇒ y→0

Hence, Z can be rewritten as-

Z =

Use the formula:

∴ Z = log a

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-

As Z =

To apply the formula we need to get the form as present in the formula. So we proceed as follows-

∵ Z =

Multiplying numerator and denominator by √(1+cos x)

⇒ Z =

Using (a+b)(a-b) = a²-b²

Z =

∵ √(1-cos²x) = sin x

⇒ Z =

{using algebra of limits}

⇒ Z =

Dividing numerator and denominator by x-

Z =

⇒ Z =

Use the formula: and

∴ Z =

{∵ log e = 1}

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential limits.

As Z =

⇒ Z =

⇒ Z =

{using properties of exponents}

⇒ Z =

{using algebra of limits}

As, x→ 5

∴ x-5→ 0

Let, y = x-5

∴ if x→5 ⇒ y→0

Hence, Z can be rewritten as-

Z =

Use the formula:

∴ Z = e⁵ log e

{∵ log e = 1}

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential limits.

As Z =

⇒ Z =

⇒ Z =

{using algebra of limits}

Use the formula:

∴ Z = e² log e

{∵ log e = 1}

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential limits.

As x→ π/2

∴ cos x → 0

Let, y = cos x

∴ if x→ π/2 ⇒ y→0

Hence, Z can be rewritten as-

Use the formula:

∴ Z = 1

{∵ log e = 1}

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z = =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential limits.

As Z =

⇒ Z =

⇒ Z =

{using algebra of limits}

Use the formula: and

∴ Z = e³ log e – 1 {∵ log e = 1}

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z = =

As we got a finite value, so no need to do any modifications.

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential and logarithmic limits.

As Z =

⇒ Z =

{Adding and subtracting 1 in numerator}

⇒ Z =

{using algebra of limits}

To get the form as present in the formula we multiply and divide 3 and 2 into both terms respectively:

⇒ Z =

Use the formula:

∴ Z = 3log e – 2log e= 3-2 = 1

{using log e = 1}

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z = =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential limits.

As, x→ 0

∴ tan x → 0

Let, y = tan x

∴ if x→ 0 ⇒ y→0

Hence, Z can be rewritten as-

Use the formula:

∴ Z = log e = 1

{∵ log e = 1}

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of variable at which the limiting value is asked, if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc)

Let Z = =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential limits.

As Z =

⇒ Z =

⇒ Z =

{using properties of exponents}

⇒ Z =

{using algebra of limits}

⇒ Z =

∴ Z =

As, x→ 0

∴ bx-sin x → 0

Let, y = bx-sin x

∴ if x→0 ⇒ y→0

Hence, Z can be rewritten as-

Z =

Use the formula:

∴ Z = log e =1

{∵ log e = 1}

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z = =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential limits.

∵ Z =

To get the desired form, we proceed as follows-

Dividing numerator and denominator by tan x-

⇒ Z =

Using algebra of limits-

Z =

Use the formula - (sandwich theorem)

∴ Z =

As, x→ 0

∴ tan x → 0

Let, y = tan x

∴ if x→ 0 ⇒ y→0

Hence, Z can be rewritten as-

Use the formula:

∴ Z = log e = 1

{∵ log e = 1}

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z = =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential limits.

As Z =

⇒ Z =

⇒ Z =

{using properties of exponents}

⇒ Z =

{using algebra of limits}

⇒ Z =

∴ Z =

As, x→ 0

∴ x-sin x → 0

Let, y = x-sin x

∴ if x→0 ⇒ y→0

Hence, Z can be rewritten as-

Z =

Use the formula:

∴ Z = log e =1

{∵ log e = 1}

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z = =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential limits.

As Z =

⇒ Z =

⇒ Z =

{using algebra of limits}

Use the formula:

∴ Z = 9 log 3

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

This question is a direct application of limits formula of exponential and logarithmic limits.

To get the desired forms, we need to include mx and nx as follows:

As Z =

⇒ Z =

{using law of exponents}

⇒ Z =

{using algebra of limits}

⇒ Z =

⇒ Z =

To get the form as present in the formula we multiply and divide by 2

∴ Z =

Use the formula:

∴ Z = 2 log a

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc)

Let Z =

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-

As Z =

As, 1-cos x = 2sin²(x/2)

∴ Z =

⇒ Z =

To get the desired form to apply the formula we need to divide numerator and denominator by x².

⇒ Z =

Using algebra of limits, we have-

Z =

Use the formula: and

∴ Z =

⇒ Z = 2 log e = 2

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z = =

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and

As Z =

⇒ Z = {using algebra of limits}

⇒ Z =

⇒ Z = {∵ sin(x-π/2) = -cos x}

As x→π/2

∴ x-π/2→0

Let x-π/2 = y and y→0

Z can be rewritten as-

Z =

Dividing numerator and denominator by sin y to get the form present in the formula

Z =

Using algebra of limits:

Z =

Use the formula: and

∴ Z =

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,0^∞ .. etc.)

Let Z =

As it is not taking any indeterminate form.

∴ Z = 0

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1^∞ .. etc.)

Let Z =

As it is taking indeterminate form.

∴ we need to take steps to remove this form so that we can get a finite value.

As, Z =

⇒ Z =

Taking log both sides-

⇒ log Z =

⇒ log Z =

{∵ log a^m = m log a}

Now it gives us a form that can be reduced to

Dividing numerator and denominator by tan²√x –

log Z =

using algebra of limits –

Let, tan²√x = y

As x→0⁺⇒ y→0⁺

∴ A =

Use the formula -

∴ A = 1

Now, B =

⇒ B =

Use the formula -

∴ B = 2

Hence,

log Z =

⇒ log_e Z = 1/2

∴ Z = e^1/2

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1^∞ .. etc.)

Let Z =

As it is taking indeterminate form-

∴ we need to take steps to remove this form so that we can get a finite value.

As, Z =

⇒ Z =

Taking log both sides-

⇒ log Z =

⇒ log Z =

{∵ log a^m = m log a}

Now it gives us a form that can be reduced to

log Z = {adding and subtracting 1 to cos x to get the form}

Dividing numerator and denominator by cos x – 1 to match with form in formula

∴ log Z =

using algebra of limits –

log Z =

∴ A =

Let, cos x - 1 = y

As x→0 ⇒ y→0

∴ A =

Use the formula -

∴ A = 1

Now, B =

∵ cos x – 1 = -2sin²(x/2) and sinx = 2sin(x/2)cos(x/2)

⇒ B =

∴ B = -cot 0 = ∞

∴ B = ∞

Hence,

log Z =

⇒ log_e Z = 0

∴ Z = e⁰ = 1

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1^∞ .. etc.)

Let Z =

As it is taking indeterminate form-

∴ we need to take steps to remove this form so that we can get a finite value.

As, Z =

⇒ Z =

Taking log both sides-

⇒ log Z =

⇒ log Z =

{∵ log a^m = m log a}

Now it gives us a form that can be reduced to

log Z =

{adding and subtracting 1 to cos x to get the form}

Dividing numerator and denominator by cos x + sin x– 1 to match with form in formula

∴ log Z =

using algebra of limits –

log Z =

∴ A =

Let, cos x + sin x - 1 = y

As x→0 ⇒ y→0

∴ A =

Use the formula -

∴ A = 1

Now, B =

∵ cos x – 1 = -2sin²(x/2) and sin x = 2sin(x/2)cos(x/2)

⇒ B =

⇒ B =

⇒ B =

Use the formula -

⇒ B =

∴ B = 1

Hence,

log Z =

⇒ log_e Z = 1

∴ Z = e¹ = e

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1^∞ .. etc.)

Let Z =

As it is taking indeterminate form-

∴ we need to take steps to remove this form so that we can get a finite value.

As, Z =

⇒ Z =

Taking log both sides-

⇒ log Z =

⇒ log Z =

{∵ log a^m = m log a}

Now it gives us a form that can be reduced to

Adding and subtracting 1 to cos x to get the form-

log Z =

Dividing numerator and denominator by cos x + a sin x– 1 to match with form in formula

∴ log Z =

using algebra of limits –

log Z =

∴ A =

Let, cos x + asin x - 1 = y

As x→0 ⇒ y→0

∴ A =

Use the formula -

∴ A = 1

Now, B =

∵ cos x – 1 = -2sin²(x/2) and sin x = 2sin(x/2)cos(x/2)

⇒ B =

⇒ B =

⇒ B =

Use the formula -

⇒ B =

∴ B = 1/a

Hence,

log Z =

⇒ log_e Z = a

∴ Z = e^a = e^a

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1^∞ .. etc.)

Let Z =

As it is taking indeterminate form-

∴ we need to take steps to remove this form so that we can get a finite value.

Z =

Take the log to bring the term in the product so that we can solve it more easily.

Taking log both sides-

log Z =

⇒ log Z =

{∵ log a^m = m log a}

⇒ log Z =

{using algebra of limits}

Still, if we put x = ∞ we get an indeterminate form,

Take the highest power of x common and try to bring x in the denominator of a term so that if we put x = ∞ term reduces to 0.

∴ log Z =

⇒ log Z =

⇒ log Z =

⇒ log Z =

∴ Log_e Z =

⇒ Z = 1/2

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1^∞ .. etc.)

Let Z =

As it is taking indeterminate form-

∴ we need to take steps to remove this form so that we can get a finite value.

Z =

Take the log to bring the power term in the product so that we can solve it more easily.

Taking log both sides-

log Z =

⇒ log Z =

{∵ log a^m = m log a}

using algebra of limits-

⇒ log Z =

⇒ log Z =

⇒ log Z =

As, 1-cos x = 2sin²(x/2)

∴ log Z =

Let (x-1)/2 = y

As x→1 ⇒ y→0

∴ Z can be rewritten as

Log Z =

⇒ log Z =

Use the formula -

∴ log Z =

⇒ log Z =

∴ Z =

Hence,

Let

Putting the limit, we get,

This is an indeterminate form, so we need to solve this limit. Taking log on both sides we get,

Now, applying L-Hospital’s rule, we get,

Applying L-hospital rule again we get,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1^∞ .. etc.)

Let Z =

As it is taking indeterminate form-

∴ we need to take steps to remove this form so that we can get a finite value.

Z =

Take the log to bring the power term in the product so that we can solve it more easily.

Taking log both sides-

log Z =

{∵ log a^m = m log a}

Now it gives us a form that can be reduced to

⇒ log Z =

Dividing numerator and denominator by to get the desired form and using algebra of limits we have-

log Z =

if we assume then as x→a ⇒ y→ 0

⇒ log Z =

Use the formula-

∴ log Z =

⇒ log Z =

Now it gives us a form that can be reduced to

Try to use it. We are basically proceeding with a hit and trial attempt.

⇒ log Z =

∵ sin (A+B) = sin A cos B + cos A sin B

⇒ log Z =

⇒ log Z=

⇒ log Z =

⇒ log Z =

Use the formula-

⇒ log Z = cot a – 0

∴ log Z = cot a

∴ Z = e^{cot a}

Hence,

As we need to find

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1^∞ .. etc.)

Let Z =

As it is taking indeterminate form-

∴ we need to take steps to remove this form so that we can get a finite value.

Z =

Take the log to bring the term in the product so that we can solve it more easily.

Taking log both sides-

log Z =

⇒ log Z =

{∵ log a^m = m log a}

⇒ log Z =

{using algebra of limits}

Still, if we put x = ∞ we get an indeterminate form,

Take highest power of x common and try to bring x in denominator of a term so that if we put x = ∞ term reduces to 0.

∴ log Z =

⇒ log Z =

⇒ log Z =

⇒ log Z =

{∵ log (3/4) is a negative value as 3/4<1}

∴ Log_e Z = -∞

⇒ Z = e^-∞ = 0

Hence,

Given limit

Putting the value of limits directly, i.e., x = 1, we have

Hence the value of the given limit is 1.

Given limit

Putting the value of limits directly, i.e. x = 0, we have

Hence the value of the given limit is 2.

Given limit

Putting the value of limits directly, i.e. x = 0, we have

Hence the value of the given limit is 0.5

Given limit

Putting the values of limits directly, i.e. x = 1, we have

Hence the value of the given limit is 3.

Given limit

Putting the values of limit directly, i.e. x = a, we have

Hence the value of the given limit is

Given limit

Putting the values of limits directly, i.e. x = 1, we have

Hence the value of the given limit is 0.5

Given limit

Putting the value of limit directly, i.e. x = 0, we have

Hence the value of the given limit is

Given the limit

Always remember the limiting value of a constant (such as 4, 13, b, etc.) is the constant itself.

So, the limiting value of constant 9 is itself, i.e., 9.

Given the limit

Putting the limiting value directly, i.e. x = 2, we have

Hence the value of the given limit is 1.

Given limit

Putting the value of limits directly, we have

Hence the value of the given limit is 6.

Given the limit

Putting the value of limits directly, i.e. x = -1, we have

Hence the value of the given limit is

Given limit

Putting the value of limit directly, i.e. x = 0, we have

Hence the value of the given limit is

Given limit

Putting the value of limits directly, i.e. x = 3, we have

Hence the value of the given limit is 0.

Given limit

Putting the value of limits directly, i.e. x = 0, we have

The given condition d ≠ 0 is reasonable because the denominator cannot be zero.

Hence the value of the given limit is .

Since the form is indeterminant

Method 1: factorization

= 2(-5)-1

= -11

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

= 4(-5) + 9

= -11

Since the form is indeterminant

Method 1: factorization

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

Since the form is indeterminant

Method 1: factorization

Since a²-b² = (a + b)(a-b)

Thus

= 3² + 3²

= 18

Method 2:

By L hospital rule:

Differentiating numerator and denominator separately:

= 54

Since the form is indeterminant

Method 1: factorization

Since a³-b³ = (a-b)(a² + b² + ab) & a²-b² = (a + b)(a-b)

= 3

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

= 3

Since the form is indeterminant

Method 1: factorization

Since a³ + b³ = (a + b)(a² + b² - ab)

= 1 + 1 + 1

= 3

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

= 12(-1/2)²

= 12/4

= 3

Since the form is indeterminant

Method 1: factorization

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

Since the form is indeterminant

Method 1: factorization

Since a²-b² = (a + b)(a-b)

= (2 + 2)(2² + 2²)

= 32

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

= 4(2)³

= 32

Since the form is indeterminant

Method 1: factorization

Method 2:

By L hospital rule:

Differentiating numerator and denominator separately:

Since the form is indeterminant

Method 1: factorization

Since a³ + b³ = (a + b)(a² + b²-ab) & a²-b² = (a + b)(a-b)

= (-1)² + (1)² - (-1)

= 3

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

= 3(-1)²

= 3

Since the form is indeterminant

Method 1: factorization

Since a³ + b³ = (a + b)(a² + b²-ab) & a²-b² = (a + b)(a-b)

=

=

= 25

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

= 25

Since the form is indeterminant

Method 1: factorization

Since a³ + b³ = (a + b)(a² + b²-ab) & a²-b² = (a + b)(a-b)

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

Since the form is

Method 1: factorization

Since a³ + b³ = (a + b)(a² + b²-ab) & a²-b² = (a + b)(a-b)

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

Since the form is

Method 1: factorization

Since a³ + b³ = (a + b)(a² + b²-ab) & a²-b² = (a + b)(a-b)

= 2

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

= 2

Since a²-b² = (a + b)(a-b)

= 2

Hence,

Since the form is indeterminant

Method 1: factorization

Since a³ + b³ = (a + b)(a² + b²-ab) & a²-b² = (a + b)(a-b)

= 2

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

= 2

Since the form is indeterminant

Method 1: factorization

Since a³ + b³ = (a + b)(a² + b²-ab) & a²-b² = (a + b)(a-b)

Since a³ + b³ = (a + b)(a² + b²-ab) & a²-b² = (a + b)(a-b)

= (√4 + 2)(4 + 4)

= (2 + 2)(4 + 4)

= 32

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

= 4(4)^3/2

= 32

Since the form is indeterminant

Method 1: factorization

Since a³ + b³ = (a + b)(a² + b²-ab) & a²-b² = (a + b)(a-b)

= 2a + 0

= 2a

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

= 2(a + 0)

= 2a

Since a³ + b³ = (a + b)(a² + b²-ab) & a²-b² = (a + b)(a-b)

= 1

Since a³ + b³ = (a + b)(a² + b²-ab) & a²-b² = (a + b)(a-b)

Since a³ + b³ = (a + b)(a² + b²-ab) & a²-b² = (a + b)(a-b)

= 6

Since the form is indeterminant

Method 1: factorization

Since a³-b³ = (a-b)(a² + b² + ab) & a²-b² = (a + b)(a-b)

=

=

=

=

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

Since the form is indeterminant

Method 1: factorization

By long division method

Since a³-b³ = (a-b)(a² + b² + ab) & a²-b² = (a + b)(a-b)

=

=

Method2: By L hospital rule:

Differentiating numerator and denominator separately:

Since the form is indeterminant

Method 1: factorization

Since a³-b³ = (a-b)(a² + b² + ab) & a²-b² = (a + b)(a-b)

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

Since the form is indeterminant

Method 1: factorization

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

Since the form is

Method 1: factorization

By long division method

Since a³-b³ = (a-b)(a² + b² + ab) & a²-b² = (a + b)(a-b)

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

Since the form is

Method 1: factorization

by dividing

Since a³-b³ = (a-b)(a² + b² + ab) & a²-b² = (a + b)(a-b)

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

= 2

Since the form is indeterminant

Method 1: factorization

Since a³-b³ = (a-b)(a² + b² + ab) & a²-b² = (a + b)(a-b)

= 1

Method 2: By L hospital rule:

Differentiating numerator and denominator separately:

= 1

Given

To find: the limit of the given equation when x tends to 0

Substituting x as 0, we get an indeterminant form of

Rationalizing the given equation

Formula: (a + b) (a - b) = a² - b²

Now we can see that the indeterminant form is removed, so substituting x as 0

We get,

Given

To find: the limit of the given equation when x tends to 0

Substituting x as 0, we get an indeterminant form of

Rationalizing the given equation

Formula: (a + b) (a - b) = a² - b²

Now we can see that the indeterminant form is removed, so substituting x as 0

We get

Given

To find: the limit of the given equation when x tends to 0

Substituting x as 0, we get an indeterminant form of

Rationalizing the given equation

Formula: (a + b) (a - b) = a² - b²

Now we can see that the indeterminant form is removed, so substituting x as 0

We get,

Given

To find: the limit of the given equation when x tends to 0

Substituting x as 0 we get an indeterminant form of

Rationalizing the given equation

Formula: (a + b) (a - b) = a² - b²