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प्रश्न
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos 2x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{x \tan x}{1 - \cos \left( 2x \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{x \tan x}{2 \sin^2 x} \right]\]
\[\text{ Dividing the numerator and thedenominator by } x^2 : \]
\[ \lim_{x \to 0} \left[ \frac{\frac{x \tan x}{x^2}}{2\frac{\sin^2 x}{x^2}} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\frac{\tan x}{x}}{2 \left( \frac{\sin x}{x} \right)^2} \right]\]
\[ = \frac{1}{2}\]
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