Evaluate the limit: lim _{x rightarrow 0}left | vedclass

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(B) Let $L = \lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)^{\frac{x}{x+1-e^x}}$.Since $\lim _{x \rightarrow 0} \frac{e^x-1}{x} = 1$ and the expo

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$e^{-1}$

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(B) Let $L = \lim _{x ightarrow 0}\left(\frac{e^x-1}{x}ight)^{\frac{x}{x+1-e^x}}$.Since $\lim _{x ightarrow 0} \frac{e^x-1}{x} = 1$ and the exponent $\frac{x}{x+1-e^x} 	o \infty$ as $x 	o 0$,this is an indeterminate form of type $1^\infty$.We use the formula $\lim _{x 	o a} f(x)^{g(x)} = e^{\lim _{x 	o a} g(x)(f(x)-1)}$.$L = e^{\lim _{x 	o 0} \left( \frac{x}{x+1-e^x} ight) \left( \frac{e^x-1}{x} - 1 ight)}$.$L = e^{\lim _{x 	o 0} \left( \frac{x}{x+1-e^x} ight) \left( \frac{e^x-1-x}{x} ight)}$.$L = e^{\lim _{x 	o 0} \frac{e^x-1-x}{x+1-e^x}}$.$L = e^{\lim _{x 	o 0} \frac{-(x+1-e^x)}{x+1-e^x}} = e^{-1}$.

Evaluate the limit: $\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)^{\frac{x}{x+1-e^x}}$

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