For the function f(x) = begin{cases} frac{e^{ | vedclass

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(A) Given $f(x) = \begin{cases} \frac{e^{1/x} - 1}{e^{1/x} + 1}, & x 
e 0 \ 0, & x = 0 \end{cases}$.First,we calculate the right-hand limit $(RHL)$

Vedclass · Accepted Answer

$\lim_{x 	o 0} f(x)$ does not exist

Vedclass · Answer

(A) Given $f(x) = \begin{cases} \frac{e^{1/x} - 1}{e^{1/x} + 1}, & x 
e 0 \ 0, & x = 0 \end{cases}$.First,we calculate the right-hand limit $(RHL)$ at $x = 0$:$\lim_{x 	o 0^+} f(x) = \lim_{h 	o 0} \frac{e^{1/h} - 1}{e^{1/h} + 1} = \lim_{h 	o 0} \frac{1 - e^{-1/h}}{1 + e^{-1/h}} = \frac{1 - 0}{1 + 0} = 1$.Next,we calculate the left-hand limit $(LHL)$ at $x = 0$:$\lim_{x 	o 0^-} f(x) = \lim_{h 	o 0} \frac{e^{-1/h} - 1}{e^{-1/h} + 1} = \frac{0 - 1}{0 + 1} = -1$.Since the left-hand limit $(-1)$ is not equal to the right-hand limit $(1)$,the limit $\lim_{x 	o 0} f(x)$ does not exist.Wait,re-evaluating the limits:For $x 	o 0^+$,$1/x 	o \infty$,so $\frac{e^{1/x}-1}{e^{1/x}+1} \approx \frac{e^{1/x}}{e^{1/x}} = 1$.For $x 	o 0^-$,$1/x 	o -\infty$,so $e^{1/x} 	o 0$,thus $\frac{0-1}{0+1} = -1$.Since $LHL 
eq RHL$,the limit does not exist. The provided option $(d)$ is incorrect based on standard calculus. However,checking the question again,if the function was defined differently,it might change. Given the provided options,the limit does not exist.

For the function $f(x) = \begin{cases} \frac{e^{1/x} - 1}{e^{1/x} + 1}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,which of the following is correct?

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