lim _{x rightarrow frac{pi}{4}} frac{8 sqrt{2 | vedclass

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Question

(A) ધારો કે $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$. આ $\frac{0}{0}$ સ્વરૂપ છે.$L'

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) ધારો કે $L = \lim _{x ightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$. આ $\frac{0}{0}$ સ્વરૂપ છે.$L'H\hat{o}pital$ નિયમ લાગુ પાડતા:$L = \lim _{x ightarrow \frac{\pi}{4}} \frac{-7(\cos x+\sin x)^{6}(-\sin x+\cos x)}{-2 \sqrt{2} \cos 2 x}$નોંધો કે $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$.આ કિંમત મૂકતા:$L = \lim _{x}$ ${ightarrow \frac{\pi}{4}} \frac{-7(\cos x+\sin x)^{6}(\cos x - \sin x)}{-2 \sqrt{2} (\cos x - \sin x)(\cos x + \sin x)}$$L = \lim _{x ightarrow \frac{\pi}{4}} \frac{7(\cos x+\sin x)^{5}}{2 \sqrt{2}}$$x = \frac{\pi}{4}$ માટે,$(\cos x + \sin x) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.$L = \frac{7(\sqrt{2})^{5}}{2 \sqrt{2}} = \frac{7 	imes 4 \sqrt{2}}{2 \sqrt{2}} = \frac{28}{2} = 14$.

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$ ની કિંમત શોધો.

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