Question:
The value of the limit $\lim_{x \to 0} \frac{e^{3x} - e^{-2x}}{\sin 4x}$ is:
The value of the limit $\lim_{x \to 0} \frac{e^{3x} - e^{-2x}}{\sin 4x}$ is:
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For $\lim_{x\to 0} \frac{e^{ax} - e^{-bx}}{\sin cx}$, the value of the limit simplifies directly to $\frac{a+b}{c}$. Here, $\frac{3 - (-2)}{4} = \frac{5}{4}$.
Updated On: Jun 3, 2026
- $\frac{5}{4}$
- $\frac{1}{4}$
- $\frac{3}{4}$
- $\frac{1}{2}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
This limit is of the indeterminate form $\frac{0}{0}$. We can resolve this using L'Hopital's Rule or standard limits.
Step 2: Meaning
Differentiating the numerator and the denominator with respect to $x$ allows us to directly evaluate the limit.
Step 3: Analysis
Applying L'Hopital's Rule: \[ \lim_{x \to 0} \frac{e^{3x} - e^{-2x}}{\sin 4x} = \lim_{x \to 0} \frac{\frac{d}{dx}\left(e^{3x} - e^{-2x}\right)}{\frac{d}{dx}(\sin 4x)} \] \[ = \lim_{x \to 0} \frac{3e^{3x} + 2e^{-2x}}{4\cos 4x} \] Substituting $x = 0$: \[ = \frac{3e^0 + 2e^0}{4\cos 0} = \frac{3(1) + 2(1)}{4(1)} = \frac{5}{4} \]
Step 4: Conclusion
The value of the limit is $\frac{5}{4}$.
Final Answer: (A)
This limit is of the indeterminate form $\frac{0}{0}$. We can resolve this using L'Hopital's Rule or standard limits.
Step 2: Meaning
Differentiating the numerator and the denominator with respect to $x$ allows us to directly evaluate the limit.
Step 3: Analysis
Applying L'Hopital's Rule: \[ \lim_{x \to 0} \frac{e^{3x} - e^{-2x}}{\sin 4x} = \lim_{x \to 0} \frac{\frac{d}{dx}\left(e^{3x} - e^{-2x}\right)}{\frac{d}{dx}(\sin 4x)} \] \[ = \lim_{x \to 0} \frac{3e^{3x} + 2e^{-2x}}{4\cos 4x} \] Substituting $x = 0$: \[ = \frac{3e^0 + 2e^0}{4\cos 0} = \frac{3(1) + 2(1)}{4(1)} = \frac{5}{4} \]
Step 4: Conclusion
The value of the limit is $\frac{5}{4}$.
Final Answer: (A)
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