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प्रश्न
\[\lim_{x \to 0} \frac{e^{2x} - e^x}{\sin 2x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{e^{2x} - e^x}{\sin \left( 2x \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{e^x \left( e^x - 1 \right)}{\sin 2x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{e^x \left( e^x - 1 \right)}{x} \times \frac{2x}{\sin 2x} \times \frac{1}{2} \right]\]
\[ = e^0 \times 1 \times \frac{1}{1} \times \frac{1}{2}\]
\[ = \frac{1}{2}\]
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