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

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(C) To check the continuity at $x = 0$,we evaluate the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the function value $f(0)$.$1$. Function va

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$f(x)$ is continuous at $x = 0$

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(C) To check the continuity at $x = 0$,we evaluate the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the function value $f(0)$.$1$. Function value: $f(0) = 0$.$2$. Left-hand limit $(LHL)$: $\lim_{x 	o 0^-} f(x) = \lim_{h 	o 0} f(0 - h) = \lim_{h 	o 0} \frac{-h}{e^{-1/h} + 1}$.As $h 	o 0^+$,$1/h 	o \infty$,so $e^{-1/h} 	o 0$.Thus,$\lim_{h 	o 0} \frac{-h}{0 + 1} = 0$.$3$. Right-hand limit $(RHL)$: $\lim_{x 	o 0^+} f(x) = \lim_{h 	o 0} f(0 + h) = \lim_{h 	o 0} \frac{h}{e^{1/h} + 1}$.As $h 	o 0^+$,$e^{1/h} 	o \infty$,so $\frac{h}{e^{1/h} + 1} 	o 0$.Since $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = 0$,the function $f(x)$ is continuous at $x = 0$.

If $f(x) = \begin{cases} \frac{x}{e^{1/x} + 1}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,then

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