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 – cos²x 

⇒ 

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 + cot²x 

⇒ 

⇒ 

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 – cot²x 

⇒

⇒ 

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