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If $f(x)=\frac{e^{1 / x}-1}{e^{1 / x}+1}$, then
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Verified Answer
The correct answer is:
$\lim _{x \rightarrow \infty} f(x)=0$
$\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow 0} \frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}$
$\therefore \lim _{x \rightarrow \infty} f(x)=0$
$\therefore \lim _{x \rightarrow \infty} f(x)=0$
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