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Class 11 NCERT Solutions- Chapter 13 Limits And Derivatives – Exercise 13.1 | Set 1
• Last Updated : 07 Apr, 2021

Question 1:

Solution:

In , as x⇢3

Put x = 3, we get

= 3+3

= 6

Question 2:

Solution:

In , as x⇢π

Put x = π, we get

Question 3:

Solution:

In , as r⇢1

Put r = 1, we get

= π

Question 4:

Solution:

In , as x⇢4

Put x = 4, we get

Question 5:

Solution:

In , as x⇢-1

Put x = -1, we get

Question 6:

Solution:

In , as x⇢0

Put x = 0, we get

As, this limit becomes undefined

Now, let’s take x+1=p and x = p-1, to make it equivalent to theorem.

As, x⇢0 ⇒ p⇢1

Here, n=5 and a = 1.

= 5(1)4

= 5

Question 7:

Solution:

In , as x⇢2

Put x = 2, we get

As, this limit becomes undefined

Now, let’s Factorise the numerator and denominator, we get

Cancelling (x-2), we have

Put x = 2, we get

Question 8:

Solution:

In , as x⇢3

Put x = 3, we get

As, this limit becomes undefined

Now, let’s Factorise the numerator and denominator, we get

Cancelling (x-3), we have

Put x = 3, we get

Question 9:

Solution:

In , as x⇢0

Put x = 0, we get

= b

Question 10:

Solution:

In , as z⇢1

Put z = 1, we get

Let’s take  = p and  = p2,

As, z⇢1 ⇒ p⇢1

Now, let’s Factorise the numerator, we get

Cancelling (p-1), we have

Put p = 1, we get

= 2

Question 11:

Solution:

In , as x⇢1

Put x = 1, we get

= 1 (As it is given a+b+c≠0)

Question 12:

Solution:

In , as x⇢-2

Firstly, lets simplify the equation

Cancelling (x+2),we get

Put x = -2, we get

Question 13:

Solution:

In , as x⇢0

Put x = 0, we get

As, this limit becomes undefined

Now, let’s multiply and divide the equation by a, to make it equivalent to theorem.

Hence, we have

=

As x⇢0, then ax⇢0

By using the theorem, we get

=

Question 14:

Solution:

In , as x⇢0

Put x = 0, we get

As, this limit becomes undefined

Now, let’s multiply and divide the numerator by ax and denominator by bx to make it equivalent to theorem.

Hence, we have

By using the theorem, we get

Question 15:

Solution:

In , as x⇢π

Put x = π, we get

As, this limit becomes undefined

Now, let’s take π-x=p

As, x⇢π ⇒ p⇢0

By using the theorem, we get

Question 16:

Solution:

In , as x⇢0

Put x = 0, we get

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