If f(x) = frac{e^{1/x} - 1}{e^{1/x} + 1},then | vedclass

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(D) To find $\lim_{x \rightarrow 0^+} f(x)$,let $x \rightarrow 0^+$,then $\frac{1}{x} \rightarrow \infty$. Thus,$\lim_{x \rightarrow 0^+} \frac{e^{1/x

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$\lim_{x \rightarrow \infty} f(x) = 0$

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(D) To find $\lim_{x \rightarrow 0^+} f(x)$,let $x \rightarrow 0^+$,then $\frac{1}{x} \rightarrow \infty$. Thus,$\lim_{x \rightarrow 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \lim_{x \rightarrow 0^+} \frac{1 - e^{-1/x}}{1 + e^{-1/x}} = \frac{1 - 0}{1 + 0} = 1$. To find $\lim_{x \rightarrow 0^-} f(x)$,let $x \rightarrow 0^-$,then $\frac{1}{x} \rightarrow -\infty$. Thus,$\lim_{x \rightarrow 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \frac{0 - 1}{0 + 1} = -1$. Since the left-hand limit and right-hand limit are not equal,$\lim_{x \rightarrow 0} f(x)$ does not exist. For $\lim_{x \rightarrow \infty} f(x)$,$\frac{1}{x} \rightarrow 0$,so $\lim_{x \rightarrow \infty} f(x) = \frac{e^0 - 1}{e^0 + 1} = \frac{1 - 1}{1 + 1} = 0$. Therefore,option $D$ is correct.

If $f(x) = \frac{e^{1/x} - 1}{e^{1/x} + 1}$,then

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