Class 11 NCERT Solutions- Chapter 13 Limits And Derivatives – Exercise 13.1 | Set 2

Question 17:

Solution:

In, as x⇢0

As we know, cos 2θ = 1-2sin²θ

Substituting the values, we get

=

Put x = 0, we get

As, this limit becomes undefined

Now, let’s multiply and divide the numerator by x² and denominator byto make it equivalent to theorem.

Hence, we have

=

=

By using the theorem, we get

=

=

=

= 4

Question 18:

Solution:

In, as x⇢0

Put x = 0, we get

As, this limit becomes undefined

Now, let’s simplify the equation to make it equivalent to theorem.

Hence, we have

=

By using the theorem, we get

=

=

Putting x=0, we have

=

Question 19:

Solution:

In, as x⇢0

Put x = 0, we get

= 0 ×1

= 0

Question 20:

Solution:

In, as x⇢0

Put x = 0, we get

As, this limit becomes undefined

Now, let’s simplify the equation to make it equivalent to theorem.

Hence, we can write the equation as follows:

=

By using the theorem, we get

=

=

=

=

Putting x=0, we have

= 1

Question 21:

Solution:

In, as x⇢0

By simplification, we get

Put x = 0, we get

As, this limit becomes undefined

Now, let’s simplify the equation to make it equivalent to theorem:

By using the trigonometric identities,

cos 2θ = 1-2sin²θ

sin 2θ = 2 sinθ cosθ

Hence, we can write the equation as follows:

=

=

Putting x=0, we have

= 0

Question 22:

Solution:

In, as x⇢

Put x =, we get

As, this limit becomes undefined

Now, let’s simplify the equation :

Let’s take

As, x⇢⇒ p⇢0

Hence, we can write the equation as follows:

=

=(As tan (π+θ) = tan θ)

=

=

Now, let’s multiply and divide the equation by 2 to make it equivalent to theorem

=

=

As p⇢0, then 2p⇢0

=

Using the theorem and putting p=0, we have

= 2×1×1

= 2

Question 23: Findand, where

Solution:

Let’s calculate, the limits when x⇢0

Here,

Left limit =

Right limit =

Limit value =

Hence,, then limit exists

Now, let’s calculate, the limits when x⇢1

Here,

Left limit =

Right limit =

Limit value =

Hence,, then limit exists

Question 24: Find, where

Solution:

Let’s calculate, the limits when x⇢1

Here,

Left limit =

Right limit =

As,

Hence, limit does not exists when x⇢1.

Question 25: Evaluate, where

Solution:

Let’s calculate, the limits when x⇢0

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Left limit =

Right limit =

As,

Hence, limit does not exists when x⇢0.

Question 26: Find , where

Solution:

Let’s calculate, the limits when x⇢0

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Left limit =

Right limit =

As,

Hence, limit does not exists when x⇢0.

Question 27: Find, where f(x)=|x|-5.

Solution:

Let’s calculate, the limits when x⇢5

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Left limit =

Right limit =

Hence,, then limit exists

Question 28: Supposeand ifwhat are possible values of a and b?

Solution:

As, it is given

Let’s calculate, the limits when x⇢1

Here,

Left limit =

Right limit =

Limit value f(1) = 4

So, as limit exists then it should satisfy

Hence, a+b = 4 and b-a = 4

Solving these equation, we get

a = 0 and b = 4

Question 29: Let a₁, a₂, . . ., a_n be fixed real numbers and define a function

f(x) = (x-a₁) (x-a₂)………… (x-an).

What is? For some a ≠ a₁, a₂, …, an, compute.

Solution:

Here, f(x) = (x-a₁) (x-a₂)………… (x-a_n).

Then,

=

= (a₁-a₁) (a₁-a₂)………… (a₁-a_n)

= 0

Now, let’s calculate for

=

= (a-a₁) (a-a₂)………… (a-a_n)

= (a-a₁) (a-a₂)………… (a-a_n)

Question 30: If

For what value (s) of a doesexists?

Solution:

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Let’s check for three cases of a:

• When a=0

Let’s calculate, the limits when x⇢0

Left limit =

Right limit =

As,

Hence, limit does not exists when x⇢0.

• When a>0

Let’s take a=2, for reference

Let’s calculate, the limits when x⇢2

Left limit =

Right limit =

As,

Hence, limit exists when x⇢2.

• When a<0

Let’s take a=-2, for reference

Let’s calculate, the limits when x⇢ -2

Left limit =

Right limit =

As,

Hence, limit exists when x⇢ -2.

Question 31: If the function f(x) satisfies, evaluate

Solution:

Here, as it is given

Put x = 1 in RHS, we get

= 2

Hence proved!

Question 32: If. For what integers m and n does bothandexists?

Solution:

Let’s calculate, the limits when x⇢0

Here,

Left limit =

Right limit =

Hence,

, then limit exists

m = n

Now, let’s calculate, the limits when x⇢1

Here,

Left limit =

Right limit =

Hence,, then limit exists.