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

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Question

(A) ધારો કે $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x}$. આ $\frac{0}{0}$ સ્વરૂપ છે.$L'H\hat{o}pital$ ના

Vedclass · Accepted Answer

$5 \sqrt{2}$

Vedclass · Answer

(A) ધારો કે $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x}$. આ $\frac{0}{0}$ સ્વરૂપ છે.$L'H\hat{o}pital$ ના નિયમનો ઉપયોગ કરતા:$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{-5(\cos x+\sin x)^4(-\sin x+\cos x)}{-2 \cos 2 x}$.$1-\sin 2x = (\cos x - \sin x)^2$ અને $\cos 2x = (\cos x - \sin x)(\cos x + \sin x)$ નિત્યસમનો ઉપયોગ કરતા:$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{5(\cos x+\sin x)^4(\cos x-\sin x)}{2(\cos x-\sin x)(\cos x+\sin x)}$.$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{5}{2}(\cos x+\sin x)^3$.$x = \frac{\pi}{4}$ મૂકતા:$L = \frac{5}{2}(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})^3 = \frac{5}{2}(\sqrt{2})^3 = \frac{5}{2}(2 \sqrt{2}) = 5 \sqrt{2}$.

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x} = $

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