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प्रश्न
\[\lim_{x \to \pi/4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}\] is equal to
विकल्प
\[\frac{1}{\sqrt{2}}\]
\[\frac{1}{2}\]
\[\frac{1}{2\sqrt{2}}\]
1
उत्तर
1/2
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}\cos x - 1}{\cot x - 1}\]
\[\text{ Rationalising the numerator, we get }: \]
\[ = \lim_{x \to \frac{\pi}{4}} \left( \frac{\sqrt{2} \cos x - 1}{\cot x - 1} \right) \times \left( \frac{\sqrt{2} \cos x + 1}{\sqrt{2} \cos x + 1} \right)\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\left( 2 \cos^2 x - 1 \right)}{\left( \cos x - \sin x \right)} \times \frac{\sin x}{\left( \sqrt{2}\cos x + 1 \right)}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\left( \cos^2 x - \sin^2 x \right)}{\left( \cos x - \sin x \right)} \times \frac{\sin x}{\left( \sqrt{2}\cos x + 1 \right)}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\left( \cos x + \sin x \right) \sin x}{\left( \sqrt{2}\cos x + 1 \right)}\]
\[ = \frac{\left( \cos \frac{\pi}{4} + \sin \frac{\pi}{4} \right) \sin \frac{\pi}{4}}{\left( \sqrt{2}\cos \frac{\pi}{4} + 1 \right)}\]
\[ = \frac{\left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right)\left( \frac{1}{\sqrt{2}} \right)}{\sqrt{2} . \frac{1}{\sqrt{2}} + 1}\]
\[ = \frac{\left( \frac{2}{\sqrt{2}} \right) \times \frac{1}{\sqrt{2}}}{2}\]
\[ = \frac{1}{2}\]
The correct answer is B.
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