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

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(C) Let $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}$.Substitute $x = \frac{\pi}{4} + h$,as $h \rightarrow

Vedclass · Accepted Answer

$\frac{3}{\sqrt{2}}$

Vedclass · Answer

(C) Let $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}$.Substitute $x = \frac{\pi}{4} + h$,as $h \rightarrow 0$.Then $\cos x + \sin x = \cos(\frac{\pi}{4} + h) + \sin(\frac{\pi}{4} + h) = \sqrt{2} \cos h$.Also,$1 - \sin 2x = 1 - \sin(2(\frac{\pi}{4} + h)) = 1 - \sin(\frac{\pi}{2} + 2h) = 1 - \cos 2h = 2 \sin^2 h$.Substituting these into the limit:$L = \lim _{h \rightarrow 0} \frac{2 \sqrt{2} - (\sqrt{2} \cos h)^3}{2 \sin^2 h} = \lim _{h \rightarrow 0} \frac{2 \sqrt{2} - 2 \sqrt{2} \cos^3 h}{2 \sin^2 h} = \lim _{h \rightarrow 0} \frac{\sqrt{2}(1 - \cos^3 h)}{\sin^2 h}$.Using $1 - \cos^3 h = (1 - \cos h)(1 + \cos h + \cos^2 h)$ and $\sin^2 h = (1 - \cos h)(1 + \cos h)$:$L = \lim _{h \rightarrow 0} \frac{\sqrt{2}(1 - \cos h)(1 + \cos h + \cos^2 h)}{(1 - \cos h)(1 + \cos h)} = \lim _{h \rightarrow 0} \frac{\sqrt{2}(1 + \cos h + \cos^2 h)}{1 + \cos h}$.As $h \rightarrow 0$,$\cos h \rightarrow 1$,so $L = \frac{\sqrt{2}(1 + 1 + 1)}{1 + 1} = \frac{3 \sqrt{2}}{2} = \frac{3}{\sqrt{2}}$.

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}=$

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