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प्रश्न
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{1 - \sqrt{2} \sin x}\]
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उत्तर
\[\lim_{x \to \frac{\pi}{4}} \left[ \frac{1 - \tan x}{1 - \sqrt{2} \sin x} \right]\]
\[\text{ It is of } \frac{0}{0} \text{ form } .\]
Rationalising the denominator, we get:
\[\lim_{x \to \frac{\pi}{4}} \left[ \frac{\left( 1 - \tan x \right) \left( 1 + \sqrt{2} \sin x \right)}{\left( 1 - \sqrt{2} \sin x \right) \left( 1 + \sqrt{2} \sin x \right)} \right]\]
\[ = \lim_{x \to \frac{\pi}{4}} \left[ \frac{\left( 1 - \tan x \right) \left( 1 + \sqrt{2} \sin x \right)}{1 - 2 \sin^2 x} \right]\]
\[ = \lim_{x \to \frac{\pi}{4}} \left[ \frac{\left( 1 - \frac{\sin x}{\cos x} \right) \left( 1 + \sqrt{2} \sin x \right)}{\cos 2x} \right]\]
\[ = \lim_{x \to \frac{\pi}{4}} \left[ \frac{\left( \cos x - \sin x \right) \left( 1 + \sqrt{2} \sin x \right)}{\cos x \cos 2x} \right] \]
\[ = \lim_{x \to \frac{\pi}{4}} \left[ \frac{\left( \cos x - \sin x \right) \left( 1 + \sqrt{2} \sin x \right)}{\cos x \cdot \left( \cos^2 x - \sin^2 x \right)} \right]\]
\[ = \lim_{x \to \frac{\pi}{4}} \left[ \frac{\left( \cos x - \sin x \right) \left( 1 + \sqrt{2} \sin x \right)}{\cos x \left[ \cos x - \sin x \right] \left[ \cos x + \sin x \right]} \right]\]
\[ = \frac{\left( 1 + \sqrt{2} \times \frac{1}{\sqrt{2}} \right)}{\left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right)}\]
\[ = \frac{2}{\frac{1}{\sqrt{2}} \times \sqrt{2}}\]
\[ = 2\]
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