$ \lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)^{\frac{x}{x+1-e^x}}= $

$ \lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)^{\frac{x}{x+1-e^x}}= $
  1. $e$
  2. $e^{-1}$
  3. $e^2$
  4. $e^{-2}$

Solution

$\begin{aligned} & \lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)^{\frac{x}{x+1-e^x}} \\ = & \lim _{x \rightarrow 0}\left[1+\frac{e^x-1}{x}-1\right]^{\frac{x}{x+1-e^x}} \\ = & \lim _{x \rightarrow 0}\left[1+\frac{e^x-1-x}{x}\right]^{\frac{-x}{e^x-1-x}} \\ = & e^{-1} \quad\left[\because \lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=e\right]\end{aligned}$

Asked in: AP EAMCET 2022 (06 Jul Shift 2)

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