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प्रश्न
\[\lim_{x \to 0} \frac{1 - \cos 2x + \tan^2 x}{x \sin x}\]
उत्तर
\[\lim_{x \to 0} \left[ \frac{1 - \cos 2x + \tan^2 x}{x \sin x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin^2 x + \tan^2 x}{x \sin x} \right] \left[ \because 1 - \cos 2A = 2 \sin^2 A \right]\]
\[\text{ Dividing numerator & denominator by } x^2 : \]
\[ \lim_{x \to 0} \left[ \frac{\frac{2 \sin^2 x}{x^2} + \frac{\tan^2 x}{x^2}}{\left( \frac{\sin x}{x} \right)} \right]\]
\[ = \frac{2 \left( 1 \right)^2 + \left( 1 \right)^2}{1} \left[ \because \lim_{x \to 0} \frac{\sin^2 x}{x^2} = 1, \lim_{x \to 0} \frac{\tan^2 x}{x^2} = 1 \right]\]
\[ = 3\]
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