mathop {lim }limits_{x to 0} frac{{{e^{1/x}} | vedclass

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Question

(D) Let $f(x) = \frac{e^{1/x} - 1}{e^{1/x} + 1}$.For the right-hand limit $(x 	o 0^+)$:$\mathop {\lim }\limits_{x 	o 0^+} \frac{e^{1/x} - 1}{e^{1/x}

Vedclass · Accepted Answer

$	ext{Does not exist}$

Vedclass · Answer

(D) Let $f(x) = \frac{e^{1/x} - 1}{e^{1/x} + 1}$.For the right-hand limit $(x 	o 0^+)$:$\mathop {\lim }\limits_{x 	o 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \mathop {\lim }\limits_{h 	o 0} \frac{e^{1/h} - 1}{e^{1/h} + 1} = \mathop {\lim }\limits_{h 	o 0} \frac{1 - e^{-1/h}}{1 + e^{-1/h}} = \frac{1 - 0}{1 + 0} = 1$.For the left-hand limit $(x 	o 0^-)$:$\mathop {\lim }\limits_{x 	o 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \mathop {\lim }\limits_{h 	o 0} \frac{e^{-1/h} - 1}{e^{-1/h} + 1} = \frac{0 - 1}{0 + 1} = -1$.Since the right-hand limit $(1)$ is not equal to the left-hand limit $(-1)$,the limit does not exist.

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{1/x}} - 1}}{{{e^{1/x}} + 1}} = $

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