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प्रश्न
\[\lim_{x \to 0} \frac{x^2 + 1 - \cos x}{x \sin x}\]
उत्तर
\[\lim_{x \to 0} \left[ \frac{x^2 + 1 - \cos x}{x \sin x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{x^2 + 2 \sin^2 \frac{x}{2}}{x \sin x} \right]\]
\[\text{ Dividing the numerator and the denominator by } x^2 : \]
\[ \lim_{x \to 0} \left[ \frac{1 + \frac{2 \sin^2 \left( \frac{x}{2} \right)}{x^2}}{\frac{\sin x}{x}} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{1 + \frac{2 \sin^2 \left( \frac{x}{2} \right)}{\left( \frac{x}{2} \right)^2 \times 4}}{\frac{\sin x}{x}} \right]\]
\[ = \frac{1 + \frac{2}{4}}{1}\]
\[ = \frac{\frac{3}{2}}{1}\]
\[ = \frac{3}{2}\]
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