A function is f(x) = begin{cases} frac{e^{1/x | vedclass

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(B) To check the continuity at $x=0$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.For $LHL$: $\lim_{x 	o 0^-} f(x) = \lim_{h

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not continuous at $x=0$

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(B) To check the continuity at $x=0$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.For $LHL$: $\lim_{x 	o 0^-} f(x) = \lim_{h 	o 0} \frac{e^{1/(0-h)}-1}{e^{1/(0-h)}+1} = \lim_{h 	o 0} \frac{e^{-1/h}-1}{e^{-1/h}+1}$. Since $e^{-1/h} 	o 0$ as $h 	o 0^+$,$LHL$ $= \frac{0-1}{0+1} = -1$.For $RHL$: $\lim_{x 	o 0^+} f(x) = \lim_{h 	o 0} \frac{e^{1/h}-1}{e^{1/h}+1} = \lim_{h 	o 0} \frac{1-e^{-1/h}}{1+e^{-1/h}} = \frac{1-0}{1+0} = 1$.Since $LHL$ $
eq$ $RHL$,the limit does not exist at $x=0$.Therefore,the function is not continuous at $x=0$.

$A$ function is $f(x) = \begin{cases} \frac{e^{1/x}-1}{e^{1/x}+1}, & \text{if } x \neq 0 \\ 0, & \text{if } x=0 \end{cases}$

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