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प्रश्न
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( \frac{\pi}{4} - x \right)^2}\]
उत्तर
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( \frac{\pi}{4} - x \right)^2}\]
\[\text{ Dividing the numerator and the denominator by }\sqrt{2}:\]
\[ \lim_{x \to \frac{\pi}{4}} \frac{1 - \left( \frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x \right)}{\frac{1}{\sqrt{2}} \left( \frac{\pi}{4} - x \right)^2}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{1 - \left( \cos \frac{\pi}{4}\cos x + \sin \frac{\pi}{4} \sin x \right)}{\frac{1}{\sqrt{2}} \left( \frac{\pi}{4} - x \right)^2}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}\left( 1 - \cos \left( \frac{\pi}{4} - x \right) \right)}{\left( \frac{\pi}{4} - x \right)^2}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \left( 2 \sin^2 \left( \frac{\frac{\pi}{4} - x}{2} \right) \right)}{\left( \frac{\pi}{4} - x \right)^2} \left[ \because 1 - \cos \theta = 2 \sin \frac{2\theta}{2} \right]\]
\[ = 2\sqrt{2} \lim_{x \to \frac{\pi}{4}} \frac{\left[ \sin^2 \left( \frac{\frac{\pi}{4} - x}{2} \right) \right]}{4 \left( \frac{\frac{\pi}{4} - x}{2} \right)^2}\]
\[ = \frac{2\sqrt{2}}{4}\]
\[ = \frac{1}{\sqrt{2}}\]
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