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

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Question

(C) ધારો કે $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}$.$x = \frac{\pi}{4} + h$ મૂકો,જ્યાં $h \rightarro

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) ધારો કે $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}$.$x = \frac{\pi}{4} + h$ મૂકો,જ્યાં $h \rightarrow 0$.તેથી $\cos x + \sin x = \cos(\frac{\pi}{4} + h) + \sin(\frac{\pi}{4} + h) = \sqrt{2} \cos h$.વળી,$1 - \sin 2x = 1 - \sin(2(\frac{\pi}{4} + h)) = 1 - \sin(\frac{\pi}{2} + 2h) = 1 - \cos 2h = 2 \sin^2 h$.આ કિંમતો લક્ષમાં મૂકતા:$L = \lim _{h \rightarrow 0} \frac{2 \sqrt{2} - (\sqrt{2} \cos h)^3}{2 \sin^2 h} = \lim _{h \rightarrow 0} \frac{2 \sqrt{2} - 2 \sqrt{2} \cos^3 h}{2 \sin^2 h} = \lim _{h \rightarrow 0} \frac{\sqrt{2}(1 - \cos^3 h)}{\sin^2 h}$.$1 - \cos^3 h = (1 - \cos h)(1 + \cos h + \cos^2 h)$ અને $\sin^2 h = (1 - \cos h)(1 + \cos h)$ નો ઉપયોગ કરતા:$L = \lim _{h \rightarrow 0} \frac{\sqrt{2}(1 - \cos h)(1 + \cos h + \cos^2 h)}{(1 - \cos h)(1 + \cos h)} = \lim _{h \rightarrow 0} \frac{\sqrt{2}(1 + \cos h + \cos^2 h)}{1 + \cos h}$.જેમ $h \rightarrow 0$,તેમ $\cos h \rightarrow 1$,તેથી $L = \frac{\sqrt{2}(1 + 1 + 1)}{1 + 1} = \frac{3 \sqrt{2}}{2} = \frac{3}{\sqrt{2}}$.

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

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