# Class 11 NCERT Solutions- Chapter 13 Limits And Derivatives – Exercise 13.1 | Set 1

Last Updated : 07 Apr, 2021

### Question 1:

Solution:

In , as xâ‡¢3

Put x = 3, we get

= 3+3

= 6

### Question 2:

Solution:

In , as xâ‡¢Ï€

Put x = Ï€, we get

### Question 3:

Solution:

In , as râ‡¢1

Put r = 1, we get

= Ï€

### Question 4:

Solution:

In , as xâ‡¢4

Put x = 4, we get

### Question 5:

Solution:

In , as xâ‡¢-1

Put x = -1, we get

### Question 6:

Solution:

In , as xâ‡¢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, xâ‡¢0 â‡’ pâ‡¢1

Here, n=5 and a = 1.

= 5(1)4

= 5

### Question 7:

Solution:

In , as xâ‡¢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 xâ‡¢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 xâ‡¢0

Put x = 0, we get

= b

### Question 10:

Solution:

In , as zâ‡¢1

Put z = 1, we get

Let’s take  = p and  = p2,

As, zâ‡¢1 â‡’ pâ‡¢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 xâ‡¢1

Put x = 1, we get

= 1 (As it is given a+b+câ‰ 0)

### Question 12:

Solution:

In , as xâ‡¢-2

Firstly, lets simplify the equation

Cancelling (x+2),we get

Put x = -2, we get

### Question 13:

Solution:

In , as xâ‡¢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 xâ‡¢0, then axâ‡¢0

By using the theorem, we get

=

### Question 14:

Solution:

In , as xâ‡¢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 xâ‡¢Ï€

Put x = Ï€, we get

As, this limit becomes undefined

Now, let’s take Ï€-x=p

As, xâ‡¢Ï€ â‡’ pâ‡¢0

By using the theorem, we get

### Question 16:

Solution:

In , as xâ‡¢0

Put x = 0, we get

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