Question:
\( \lim_{x \to 0} \frac{e^{\sin x} - 1}{x} \) is equal to
\( \lim_{x \to 0} \frac{e^{\sin x} - 1}{x} \) is equal to
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For $x \to 0$, use $\sin x \approx x$ and $e^x \approx 1 + x$.
Updated On: Apr 23, 2026
- $0$
- $e$
- $1$
- Does not exist
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The Correct Option is C
Solution and Explanation
Concept:
Use standard limits and small angle approximation.
Step 1: Use expansion of exponential.
\[ e^{\sin x} \approx 1 + \sin x \]
Step 2: Substitute in limit.
\[ \frac{e^{\sin x} - 1}{x} \approx \frac{\sin x}{x} \]
Step 3: Use standard limit.
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \] Conclusion:
Limit = 1
Step 1: Use expansion of exponential.
\[ e^{\sin x} \approx 1 + \sin x \]
Step 2: Substitute in limit.
\[ \frac{e^{\sin x} - 1}{x} \approx \frac{\sin x}{x} \]
Step 3: Use standard limit.
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \] Conclusion:
Limit = 1
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