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$$
\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)^{\frac{x}{x+1-e^x}}=
$$
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\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)^{\frac{x}{x+1-e^x}}=
$$
Solution:
1321 Upvotes
Verified Answer
The correct answer is:
$e^{-1}$
$\begin{aligned} & \lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)^{\frac{x}{x+1-e^x}} \\ = & \lim _{x \rightarrow 0}\left[1+\frac{e^x-1}{x}-1\right]^{\frac{x}{x+1-e^x}} \\ = & \lim _{x \rightarrow 0}\left[1+\frac{e^x-1-x}{x}\right]^{\frac{-x}{e^x-1-x}} \\ = & e^{-1} \quad\left[\because \lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=e\right]\end{aligned}$
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