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प्रश्न
\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to
विकल्प
\[5\sqrt{2}\]
\[3\sqrt{2}\]
\[\sqrt{2}\]
none of these
उत्तर
\[ = \lim_{x \to \frac{\pi}{4}} \frac{2^\frac{5}{2} - \left( \left( \cos x + \sin x \right)^2 \right)^\frac{5}{2}}{2 - \left( 1 + \sin 2x \right)}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{2^\frac{5}{2} - \left( \left( \cos x + \sin x \right)^2 \right)^\frac{5}{2}}{2 - \left( \cos x + \sin x \right)^2}\]
\[Let t = \left( \cos x + \sin x \right)^2 \]
\[ x \to \frac{\pi}{4}\]
\[ \therefore t = \left( \cos \frac{\pi}{4} + \sin \frac{\pi}{4} \right)^2 \to \left( \sqrt{2} \right)^2 = 2\]
\[ = \lim_{t \to 2} \frac{2^\frac{5}{2} - \left( t \right)^\frac{5}{2}}{2 - \left( t \right)}\]
\[ = \frac{5}{2} \left( 2 \right)^\frac{3}{2} \left[ \because \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n - 1} \right]\]
\[ = 5\sqrt{2}\]
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