# 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  My Personal Notes arrow_drop_up