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

### Evaluate the following limits in Exercises 1 to 22.

### 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 = p

^{2},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

=