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