lim _{x rightarrow frac{pi}{4}} frac{sqrt{2} | vedclass

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(C) Given limit: $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}$ Substituting $x = \frac{\pi}{4}$,we get the form $\frac{

Vedclass · Accepted Answer

$1/2$

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(C) Given limit: $L = \lim _{x ightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}$ Substituting $x = \frac{\pi}{4}$,we get the form $\frac{\sqrt{2}(\frac{1}{\sqrt{2}})-1}{1-1} = \frac{0}{0}$. Applying $L'	ext{Hospital rule}$ by differentiating the numerator and denominator with respect to $x$: $L = \lim _{x ightarrow \frac{\pi}{4}} \frac{\frac{d}{dx}(\sqrt{2} \cos x - 1)}{\frac{d}{dx}(\cot x - 1)}$ $L = \lim _{x ightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \sin x}{-\operatorname{cosec}^2 x} = \lim _{x ightarrow \frac{\pi}{4}} \frac{\sqrt{2} \sin x}{\operatorname{cosec}^2 x}$ $L = \lim _{x ightarrow \frac{\pi}{4}} \sqrt{2} \sin x \sin^2 x = \lim _{x ightarrow \frac{\pi}{4}} \sqrt{2} \sin^3 x$ $L = \sqrt{2} 	imes (\frac{1}{\sqrt{2}})^3 = \sqrt{2} 	imes \frac{1}{2\sqrt{2}} = \frac{1}{2}$.

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}$ is equal to

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