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प्रश्न
\[\lim_{x \to 0} \frac{x\left( e^x - 1 \right)}{1 - \cos x}\]
उत्तर
\[\lim_{x \to 0} \left[ \frac{x \left( e^x - 1 \right)}{1 - \cos x} \right]\]
Dividing the numerator and the denominator by x2:
\[= \lim_{x \to 0} \left[ \frac{\left( \frac{e^x - 1}{x} \right)}{\left( \frac{1 - \cos x}{x^2} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{e^x - 1}{x} \times \frac{1}{\frac{2 \sin^2 \left( \frac{x}{2} \right)}{4 \times \frac{x^2}{4}}} \right]\]
\[ = 1 \times \frac{2}{1^2}\]
\[ = 2\]
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