Limits

To evaluate:

Formula used:

We have,

As , we have

Thus, the value of 3.

To evaluate:

Formula used:

We have,

As , we have

1²-4(1)+3

0

Thus, the value of 0.

To evaluate:

Formula used:

We have,

As , we have

=3

Thus, the value of 3.

To evaluate:

Formula used:

We have,

As , we have

Thus, the value of -3.

To evaluate:

Formula used:

We have,

As , we have

Thus, the value of -10.

To evaluate:

Formula used:

We have,

and

x³-y³=(x-y)(x²+xy+y²)

As , we have

Thus, the value of3.

To evaluate:

Formula used:

We have,

and

x³+y³=(x+y)(x²-xy+y²)

As , we have

4-4+4

4

Thus, the value of4

To evaluate:

Formula used:

We have,

and

As , we have

= 18 × 6

= 108

Thus, the value of108.

To evaluate:

Formula used:

We have,

and

As , we have

Thus, the value of.

To evaluate:

Formula used:

We have,

As , we have

Thus, the value of 2.

To evaluate:

Formula used:

We have,

and

x³-y³=(x-y)(x²+xy+y²)

As , we have

Thus, the value ofis 6.

To evaluate:

Formula used:

We have,

=my^m-1

As , we have

Thus, the value ofis

To evaluate:

Formula used:

We have,

=my^m-1

As , we have

Thus, the value ofis

To evaluate:

Formula used:

We have,

=my^m-1

As , we have

Thus, the value ofis

To evaluate:

Formula used:

We have,

=my^m-1

As , we have

Thus, the value ofis n.

To evaluate:

Formula used:

We have,

=my^m-1

As , we have

Thus, the value ofis

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis 0.

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis

To evaluate:

Formula used:

We have,

As , we have

Thus, the value of .

To evaluate:

Formula used:

Multiplying numerator and denominator by

Thus, the value ofis

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis -8.

To evaluate:

Formula used:

Multiplying numerator and denominator with conjugates of numerator and denominator i.e

Thus, the value ofis

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

-1

Thus, the value ofis -1.

To Evaluate:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

Therefore,

Hence,

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis 4.

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis e².

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis e⁴.

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

3-2

1

Thus, the value ofis 1.

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

1-1

0

Thus, the value ofis 0.

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

b-a

Thus, the value ofis b-a.

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis.

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis.

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis

To evaluate:

Formula used:

L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

then

As , we have

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value ofis

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfied any one from 7 indeterminate forms.

In this Case, indeterminate Form is

Formula used: = 1

So = ) = =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfied any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = 1

So = ) = =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfied any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = 1

So = ) = =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfied any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = 1 ‘

So = ) = =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = 1 and = 1

So = ) = =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = 1 and = 1

So = ) = =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = 1 and = 1

So = ) = =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, inderterminate Form is

Formula used: = 1

So = - + ) = - + )

By using the above formula, we have

- + ) = 1 - 2 + 5 = 0

Therefore, = 0

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = 1 or we can used L hospital Rule,

So, by using the rule, Differentiate numerator and denominator

= = = -3

Therefore, = -3

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = 1 or we can used L hospital Rule,

So, by using the above formula, we have

Divide numerator and denominator by x,

= = = = = 1

ALTER:by using the rule, Differentiate numerator and denominator

= = 1

Therefore, = 1

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = 1 or we can used L hospital Rule,

So, by using the above formula, we have

Divide numerator and denominator by x,

= = = = =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = 1 or we can used L hospital Rule,

So, by using the above formula, we have

Divide numerator and denominator by x,

= = = = = -2

Therefore, = -2

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = 1 and = 1

So, by using the above formula, we have

Divide numerator and denominator by x,

= = = = 2

Therefore, = 2

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

NOTE : tan x – sin x = – sin x = = sin x ()

= =

Divide numerator and denominator by x²,

=

Formula used: = 1/2 and = 1 or we can used L hospital Rule,

So, by using the above formula, we have

= =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form are 0

Formula used: = 1

So, by using the above formula, we have

= = 1

Therefore, = 1

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is 0

Formula used: = 1

So, by using the above formula, we have

= =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = 1

So, by using the above formula, we have

= =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = 1

So, by using the above formula, we have

= =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = 1

So, by using the above formula, we have

= = [Divide and multiply with 2 on denominator]

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

[NOTE: 1 – cos x = 2 sin²(x/2)]

Formula used: = 1

So, by using the above formula, we have

=

Divide numerator and denominator by x², we have

= = = = =

[NOTE: = ]

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: =

So, by using the above formula, we have

= =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = and = 1

Divide numerator and denominator by x², we have

So, by using the above formula, we have

= =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: = and = 1

Divide numerator and denominator by x², we have

So, by using the above formula, we have

= = =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: =

Divide numerator and denominator by x², we have

So, by using the above formula, we have

= = = =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

Formula used: =

Divide numerator and denominator by m² and n², we have

So, by using the above formula, we have

= =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

We know that sin 2x = 2 sin x cosx

Formula used: = and =

So, by using the above formula, we have

= = = = = 1

Therefore, = 1

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

NOTE : tan x – sin x = – sin x = = sin x ()

= =

Formula used: = 1/2 and = 1 or we can used L hospital Rule,

So, by using the above formula, we have

=

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

= = = = 4

Therefore,=

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

cosec x – cot x = (1 – cos x)/sinx

= = =

Formula used: = 1/2 and = 1

= =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form are

cosec 2x – cot 2x = (1 – cos 2x)/sin 2x

= = =

Formula used: = 1/2 and = 1

= = =

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

= =

Formula used: = 1/2 and = 1

= 4

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

By using L hospital Rule,

Differtiate both sides w.r.t x

So = So = = = 2

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

By using L hospital Rule,

Differtiate both sides w.r.t x

So = = = = 2

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Formis

By using L hospital Rule,

Differtiate both sides w.r.t x

So = = = -2

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

By using L hospital Rule,

Differtiate both sides w.r.t x

So =

Again, indeterminate Form is

So, Differtiate both sides w.r.t x again, we have

=

Again, indeterminate Form is

So, Differtiate both sides w.r.t x again, we have

= = = = - 4

Therefore, =

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is

By using L hospital Rule,

Differtiate both sides w.r.t x

So = = =

Again, indeterminate Form is

So, Differtiate both sides w.r.t x again, we have

= = = = =

Therefore, =

[]

= -sin (a)

=cosa

[Multiply and divide by √x-√a]

=2√a×cosa

=2√acosa

= 2 × 1

=8

= 4

=2cos2

=1

=2

=2a²cosa+2asina

=-1+e³

As x tends to 0, tan(x) also tends to zero,

So,

=1

=1×1

=1

=1

As x tends to, tends to zero.

Let,

=2

=4

=0

=1

=1

=sin³a

Let,

=4

Let,

Let,

= 2

= -3

Left Hand Limit(L.H.L.):

= - ( 3 - 3)

= 0

Right Hand Limit(R.H.L.):

= 3-3

= 0

Since,

We can say that the limit exists and

Left Hand Limit(L.H.L.):

= -1

Right Hand Limit(R.H.L.):

=1

Since does not exist

Left Hand Limit(L.H.L.):

= -1

Right Hand Limit(R.H.L.):

= 1

Thus, does not exist.

Left Hand Limit(L.H.L.):

= 1 + (1)²

= 1 + 1

= 2

Right Hand Limit(R.H.L.):

= 2 – (1)

= 2 – 1

=1

Thus, does not exist.

Left Hand Limit(L.H.L.):

= 2

Right Hand Limit(R.H.L.):

= 0

Thus, does not exist.

Left Hand Limit(L.H.L.):

= 5(1) – 4

= 5 – 4

= 1

Right Hand Limit(R.H.L.):

= 4 (1)³ – 3 (1)

= 4 – 3

= 1

Thus, = 1

Left Hand Limit(L.H.L.):

= 4 (2) – 5

= 8 – 5

= 3

Right Hand Limit(R.H.L.):

= 2 – a

Since it exists,

→ 3 = 2 – a

→ a = 2 – 3

→ a = -1

Left Hand Limit(L.H.L.):

= 3

Right Hand Limit(R.H.L.):

= 1

Since

Thus, does not exist.

Left Hand Limit(L.H.L.):

= 0 + k

= k

Right Hand Limit(R.H.L.):

= cos (0)

= 1

It is given that exists. Therefore,

→ k = 1

Let x = 0+h for x tending to 0⁺

Since x→ 0, h also tends to 0

Right Hand Limit(R.H.L.):

=

=

Let x=0 -h for x tending to 0^-

Since x→0, h also tends to 0.

Left Hand Limit(L.H.L.):

=

= - ∞

Since,

Thus, does not exist.

Let x = 0 + h, when x is tends to 0⁺

Since x tends to 0, h will also tend to 0.

Right Hand Limit(R.H.L):

=

=∞

Let x = 0 - h, when x is tends to 0^-

Since x tends to 0, h will also tend to 0.

Left Hand Limit(L.H.L.):

=

= ∞

Thus,

∴ .

Left Hand Limit(L.H.L.):

=

= e^∞

Right Hand Limit(R.H.L.):

= e^-∞

=

[ Formula , anything to the power infinity is also infinity. Thus ]

=0

Since

∴ does not exist.

Let x = 0 + h, when x is tends to 0⁺

Since x tends to 0, h will also tend to 0.

Right Hand Limit(R.H.L.):

=

= sin ∞

= ∞

Let x = 0 - h, when x is tends to 0^-

Since x tends to 0, h will also tend to 0.

Left Hand Limit(L.H.L.):

=

=

= - sin ∞

= -∞

Since,

∴ does not exist.

Left Hand Limit(L.H.L.):

= -1

Right Hand Limit(R.H.L.):

= 1

Since

Thus, does not exist.

Let

→

or,

or , h → 0

Putting this in the original sum,

[ Applying formula ]

= -k × 1

= -k

It is given that

∴ -k = 3

→ k = -3