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प्रश्न
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( 4x - \pi \right)^2}\]
उत्तर
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( 4x - \pi \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{2} - \left\{ \cos \left( \frac{\pi}{4} + h \right) + \sin \left( \frac{\pi}{4} + h \right) \right\}}{\left( 4\left( \frac{\pi}{4} + h \right) - \pi \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{2} - \left\{ \cos \frac{\pi}{4} \cos h - \sin \frac{\pi}{4} \sin h + \sin \frac{\pi}{4} \cos h + \cos \frac{\pi}{4} \sin h \right\}}{\left( 4h \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{2} - \sqrt{2}\cos h}{\left( 4h \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{2} \left( 1 - \cos h \right)}{16 h^2}\]
\[ = \lim_{h \to 0} \frac{2\sqrt{2} \sin^2 \frac{h}{2}}{64 \times \frac{h^2}{4}}\]
\[ = \frac{\sqrt{2}}{32} \times \left( 1 \right)^2 \]
\[ = \frac{1}{16\sqrt{2}}\]
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