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Question: Answered & Verified by Expert
$\lim _{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$ is equal to
MathematicsLimitsJEE MainJEE Main 2024 (09 Apr Shift 2)
Options:
  • A 0
  • B $\frac{-2}{e}$
  • C e
  • D $e-e^2$
Solution:
1256 Upvotes Verified Answer
The correct answer is: e
$\begin{aligned} & \operatorname{Lim}_{x \rightarrow 0} \frac{e-e^{\frac{1}{2 x} \ln (1+2 x)}}{x} \\ & =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\left(e^{\frac{\ln (1+2 x)}{2 x}-1}-1\right)}{x} \\ & =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\ln (1+2 x)-2 x}{2 x^2} \\ & =(-e) \times(-1) \frac{4}{2 \times 2}=e\end{aligned}$

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