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

Last Updated : 07 Apr, 2021
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Question 1. \lim_{x \to \pi}\frac{1+\cos x}{\tan ^2x}

Solution:

Given, \lim_{x \to \pi}\frac{1+\cos x}{\tan ^2x}

By Applying limits, we get,

⇒ \frac{1+\cos \pi}{\tan ^2\pi} = \frac{0}{0}   (Indeterminate form or 0/0 form)

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

On substituting \tan ^2x=\frac{sin^ 2x}{cos ^2x}  we get,

⇒  \lim_{x \to \pi}\frac{(1+\cos x)(\cos ^2x)}{\sin ^2x}

We know, sin2x + cos2x = 1 

⇒ sin2x = 1 – cos2x  

⇒  \lim_{x \to \pi}\frac{(1+\cos x)(\cos ^2x)}{1-\cos ^2x}

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

⇒  \lim_{x \to \pi}\frac{(1+\cos x)(\cos ^2x)}{(1+\cos x)(1-\cos x)}

⇒  \lim_{x \to \pi}\frac{\cos ^2x}{1-\cos x}

Applying limits we get,

⇒  \frac{\cos ^2\pi}{1-\cos \pi}

⇒ \frac{(-1)^2}{1-(-1)}=\frac{1}{2}

Therefore, the value of  \lim_{x \to \pi}\frac{1+\cos x}{\tan ^2x} =\frac{1}{2}

Question 2. \lim_{x \to \frac{\pi}{4}}\frac{\cosec ^2x-2}{\cot x-1}

Solution: 

Given, \lim_{x \to \frac{\pi}{4}}\frac{\cosec ^2x-2}{\cot x-1}

Applying the limits, we get,

⇒ \frac{\cosec ^2\frac{\pi}{4}-2}{\cot \frac{\pi}{4}-1}=\frac{2-2}{1-1}=\frac{0}{0}  (Indeterminate form)

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

We know, cosec2x − cot2x = 1  

⇒ cosec2x = 1 + cot2x  

⇒  \lim_{x \to \frac{\pi}{4}}\frac{1+\cot ^2x-2}{\cot x-1}

⇒  \lim_{x \to \frac{\pi}{4}}\frac{\cot ^2x-1}{\cot x-1}

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

⇒  \lim_{x \to \frac{\pi}{4}}\frac{(\cot x-1)(\cot x+1)}{\cot x-1}

⇒  \lim_{x \to \frac{\pi}{4}}\cot x+1

Applying the limits, we get, 

⇒ \cot \frac{\pi}{4}+1 =2

Therefore, the value of  \lim_{x \to \frac{\pi}{4}}\frac{\cosec ^2x-2}{\cot x-1} =2

Question 3.\lim_{x \to \frac{\pi}{6}}\frac{\cot ^2x-3}{\cosec x-2}

Solution: 

Given, \lim_{x \to \frac{\pi}{6}}\frac{\cot ^2x-3}{\cosec x-2}

Applying the limits, we get,

⇒ \frac{\cot ^2\frac{\pi}{6}-3}{\cosec \frac{\pi}{6}-2}=\frac{3-3}{2-2}=\frac{0}{0}  (Indeterminate form)

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

⇒  \lim_{x \to \frac{\pi}{6}}\frac{\cosec ^2x-1-3}{\cosec x-2}

⇒  \lim_{x \to \frac{\pi}{6}}\frac{\cosec ^2x-4}{\cosec x-2}

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

 â‡’  \lim_{x \to \frac{\pi}{6}}\frac{(\cosec x-2)(\cosec x+2)}{\cosec x-2}

⇒  \lim_{x \to \frac{\pi}{6}} (\cosec x+2)

Applying the limits, we get,

⇒ \cosec \frac{\pi}{6}+2=4

Therefore, the value of  \lim_{x \to \frac{\pi}{6}}\frac{\cot ^2x-3}{\cosec x-2} =4

Question 4. \lim_{x \to \frac{\pi}{4}}\frac{2-\cosec ^2x}{1-\cot x}

Solution:

Given, \lim_{x \to \frac{\pi}{4}}\frac{2-\cosec ^2x}{1-\cot x}

Applying the limits we get,

⇒ \frac{2-\cosec ^2\frac{\pi}{4}}{1-\cot \frac{\pi}{4}}=\frac{2-2}{1-1}=\frac{0}{0}   (Indeterminate form)

So, we cannot just apply the limits.

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

⇒ \lim_{x \to \frac{\pi}{4}}\frac{2-(1+\cot ^2x)}{1-\cot x}

⇒  \lim_{x \to \frac{\pi}{4}}\frac{1-\cot ^2x}{1-\cot x}

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

⇒ \lim_{x \to \frac{\pi}{4}}\frac{(1-\cot x)(1+\cot x)}{1-\cot x}

⇒  \lim_{x \to \frac{\pi}{4}} (1+\cot x)

Applying the limits we get,

⇒ (1+\cot \frac{\pi}{4})=2

Therefore, The value of  \lim_{x \to \frac{\pi}{4}}\frac{2-\cosec ^2x}{1-\cot x} =2

Question 5.\lim_{x \to \pi}\frac{\sqrt{2+\cos x}-1}{(\pi -x)^2}

Solution:

Given, \lim_{x \to \pi}\frac{\sqrt{2+\cos x}-1}{(\pi -x)^2}

Applying the limits, we get,

⇒ \frac{\sqrt{2+\cos \pi}-1}{(\pi -\pi)^2}=\frac{0}{0} (Indeterminate form)

So, we cannot just apply the limits.

Rationalizing the numerator(multiplying and dividing with \sqrt{2+\cos x}+1 )

⇒  \lim_{x \to \pi}\frac{(\sqrt{2+\cos x}-1)(\sqrt{2+\cos x}+1)}{(\pi -x)^2(\sqrt{2+\cos x}+1)}

⇒  \lim_{x \to \pi}\frac{(2+\cos x-1)}{(\pi -x)^2(\sqrt{2+\cos x}+1)}

Let x = Ï€ − h

If x → Ï€, h → 0  

Substituting x = Ï€ − h we get,

⇒ \lim_{h \to 0}\frac{(1+\cos (\pi-h))}{(\pi -(\pi-h))^2(\sqrt{2+\cos (\pi-h)}+1)}

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

⇒  \lim_{h \to 0}\frac{(1-\cos h)}{h^2(\sqrt{2-\cos h}+1)}

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

⇒ \lim_{h \to 0}\frac{2\sin^2(\frac{h}{2})}{(4\times\frac{h^2}{4})(\sqrt{2-\cos h}+1)}

⇒ \frac{1}{2} \lim_{h \to 0}[(\frac{\sin (\frac{h}{2})}{(\frac{h}{2})})^2\times(\frac{1}{\sqrt{2-\cos h}+1)})]

We know that, \lim_{x \to 0}\frac{\sin x}{x}=1

Applying the limits, we get, 

⇒ \frac{1}{2}\times1\times\frac{1}{\sqrt{2-1}+1}

⇒ 1/2 x 1/2 = 1/4 

Therefore, the value of  \lim_{x \to \pi}\frac{\sqrt{2+\cos x}-1}{(\pi -x)^2} =\frac{1}{4}

Question 6.\lim_{x \to \frac{3\pi}{2}}\frac{1+\cosec ^3x}{\cot^ 2x}

Solution:

Given, \lim_{x \to \frac{3\pi}{2}}\frac{1+\cosec ^3x}{\cot^ 2x}

Applying the limits, we get,

⇒ \frac{1+\cosec ^3(\frac{3\pi}{2})}{\cot^ 2(\frac{3\pi}{2})}=\frac{0}{0} (Indeterminate form)

So, we cannot just directly apply the limits,

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

⇒ \lim_{x \to \frac{3\pi}{2}}\frac{(1+\cosec x)(1-\cosec x+\cosec^ 2x)}{\cosec^ 2x-1}

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

⇒  \lim_{x \to \frac{3\pi}{2}}\frac{(1+\cosec x)(1-\cosec x+\cosec^ 2x)}{(\cosec x-1)(\cosec x+1)}

⇒  \lim_{x \to \frac{3\pi}{2}}\frac{(1-\cosec x+\cosec^ 2x)}{\cosec x-1}

Applying the limits, we get,

⇒ \frac{(1-\cosec (\frac{3\pi}{2})+\cosec^ 2(\frac{3\pi}{2}))}{\cosec (\frac{3\pi}{2})-1} =\frac{3}{-2}

Therefore, the value of  \lim_{x \to \frac{3\pi}{2}}\frac{1+\cosec ^3x}{\cot^ 2x} =\frac{-3}{2}



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