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

### Question 1.

Solution:

Given,

By Applying limits, we get,

â‡’ =    (Indeterminate form or 0/0 form)

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

On substituting  we get,

â‡’

We know, sin2x + cos2x = 1

â‡’ sin2x = 1 – cos2

â‡’

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

â‡’

â‡’

Applying limits we get,

â‡’

â‡’

Therefore, the value of

### Question 2.

Solution:

Given,

Applying the limits, we get,

â‡’  (Indeterminate form)

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

We know, cosec2x âˆ’ cot2x = 1

â‡’ cosec2x = 1 + cot2

â‡’

â‡’

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

â‡’

â‡’

Applying the limits, we get,

â‡’

Therefore, the value of

### Question 3.

Solution:

Given,

Applying the limits, we get,

â‡’  (Indeterminate form)

We know, cosec2x âˆ’ cot2x = 1 â‡’ cot2x = cosec2x – 1

â‡’

â‡’

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

â‡’

â‡’

Applying the limits, we get,

â‡’

Therefore, the value of

### Question 4.

Solution:

Given,

Applying the limits we get,

â‡’   (Indeterminate form)

So, we cannot just apply the limits.

We know, cosec2x âˆ’ cot2x = 1 â‡’ cosec2x = 1 – cot2

â‡’

â‡’

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

â‡’

â‡’

Applying the limits we get,

â‡’

Therefore, The value of

### Question 5.

Solution:

Given,

Applying the limits, we get,

â‡’ (Indeterminate form)

So, we cannot just apply the limits.

Rationalizing the numerator(multiplying and dividing with )

â‡’

â‡’

Let x = Ï€ âˆ’ h

If x â†’ Ï€, h â†’ 0

Substituting x = Ï€ âˆ’ h we get,

â‡’

We know that cos(Ï€ âˆ’ x) = âˆ’cosx substituting we get,

â‡’

By using cos2x = 1 âˆ’ 2sin2x â‡’ cos h = 1 âˆ’ 2sin2(hâ€‹/2)

â‡’

â‡’

We know that,

Applying the limits, we get,

â‡’

â‡’ 1/2 x 1/2 = 1/4

Therefore, the value of

### Question 6.

Solution:

Given,

Applying the limits, we get,

â‡’ (Indeterminate form)

So, we cannot just directly apply the limits,

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

â‡’

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

â‡’

â‡’

Applying the limits, we get,

â‡’

Therefore, the value of

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