The limiting value of the function f(x) = fra | vedclass

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(D) Let $x = \frac{\pi}{4} + h$. As $x 	o \frac{\pi}{4}$,$h 	o 0$. Then $\cos x + \sin x = \cos(\frac{\pi}{4} + h) + \sin(\frac{\pi}{4} + h) = \sqrt

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

Vedclass · Answer

(D) Let $x = \frac{\pi}{4} + h$. As $x 	o \frac{\pi}{4}$,$h 	o 0$. Then $\cos x + \sin x = \cos(\frac{\pi}{4} + h) + \sin(\frac{\pi}{4} + h) = \sqrt{2} \cos h$. Also,$1 - \sin 2x = 1 - \sin(\frac{\pi}{2} + 2h) = 1 - \cos 2h = 2 \sin^2 h$. The expression becomes $\lim_{h 	o 0} \frac{2\sqrt{2} - (\sqrt{2} \cos h)^3}{2 \sin^2 h} = \lim_{h 	o 0} \frac{2\sqrt{2}(1 - \cos^3 h)}{2 \sin^2 h} = \sqrt{2} \lim_{h 	o 0} \frac{(1 - \cos h)(1 + \cos h + \cos^2 h)}{\sin^2 h}$. Using $\lim_{h 	o 0} \frac{1 - \cos h}{\sin^2 h} = \lim_{h 	o 0} \frac{1 - \cos h}{1 - \cos^2 h} = \lim_{h 	o 0} \frac{1}{1 + \cos h} = \frac{1}{2}$. Thus,the limit is $\sqrt{2} 	imes \frac{1}{2} 	imes (1 + 1 + 1) = \sqrt{2} 	imes \frac{3}{2} = \frac{3}{\sqrt{2}}$.

The limiting value of the function $f(x) = \frac{2\sqrt{2} - (\cos x + \sin x)^3}{1 - \sin 2x}$ as $x \to \frac{\pi}{4}$ is

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