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