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• RD Sharma Class 11 Solutions for Maths

# Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.9

### Question 1.

Solution:

Given,

By Applying limits, we get,

=    (Indeterminate form or 0/0 form)

So, we cannot just directly apply the limits as we got indeterminate form.

On substituting  we get,

⇒

We know, sin2x + cos2x = 1

⇒ sin2x = 1 – cos2

⇒

By using a2 b2 = (a + b)(a b) we get,

⇒

⇒

Applying limits we get,

⇒

⇒

Therefore, the value of

### Question 2.

Solution:

Given,

Applying the limits, we get,

⇒  (Indeterminate form)

So, we cannot just directly apply the limits as we got indeterminate form.

We know, cosec2x − cot2x = 1

⇒ cosec2x = 1 + cot2

⇒

⇒

By using formula, a2 b2 = (a + b)(a b) we get,

⇒

⇒

Applying the limits, we get,

⇒

Therefore, the value of

### Question 3.

Solution:

Given,

Applying the limits, we get,

⇒  (Indeterminate form)

We know, cosec2x − cot2x = 1 ⇒ cot2x = cosec2x – 1

⇒

⇒

By using formula, a2 b2 = (a + b)(a b) we get,

⇒

⇒

Applying the limits, we get,

⇒

Therefore, the value of

### Question 4.

Solution:

Given,

Applying the limits we get,

⇒   (Indeterminate form)

So, we cannot just apply the limits.

We know, cosec2x − cot2x = 1 ⇒ cosec2x = 1 – cot2

⇒

By using formula, a2 b2 = (a + b)(a b) we get,

⇒

Applying the limits we get,

⇒

Therefore, The value of

### Question 5.

Solution:

Given,

Applying the limits, we get,

⇒ (Indeterminate form)

So, we cannot just apply the limits.

Rationalizing the numerator(multiplying and dividing with )

⇒

⇒

Let x = π − h

If x π, h → 0

Substituting x = π − h we get,

We know that cos(π x) = −cosx substituting we get,

⇒

By using cos2x = 1 − 2sin2x cos h = 1 − 2sin2(h​/2)

⇒

We know that,

Applying the limits, we get,

⇒

⇒ 1/2 x 1/2 = 1/4

Therefore, the value of

### Question 6.

Solution:

Given,

Applying the limits, we get,

⇒ (Indeterminate form)

So, we cannot just directly apply the limits,

By using the formula, a3 + b3 = (a + b)(a2 ab + b2) we get,

By using formula, a2 b2 = (a + b)(a b)

⇒

⇒

Applying the limits, we get,

⇒

Therefore, the value of

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