AP EAMCET · Maths · Limits
\[
\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)^{\frac{x}{x+1-e^x}}=
\]
- A \(e\)
- B \(e^{-1}\)
- C \(e^2\)
- D \(e^{-2}\)
Answer & Solution
Correct Answer
(B) \(e^{-1}\)
Step-by-step Solution
Detailed explanation
\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}…
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