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If $f(x)=\left\{\begin{array}{r}\frac{x}{e^{1 / x}+1}, \text { when } x \neq 0 \\ 0, \text { when } x=0\end{array}\right.$ then
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Verified Answer
The correct answer is:
$f(x)$ is continuous at $x=0$
$f(0)=0$
$f(0-)=\lim _{h \rightarrow 0} \frac{-h}{e^{-1 / h}+1}=\lim _{h \rightarrow 0} \frac{-h}{1+\frac{1}{e^{1 / h}}}=0$
$f(0+)=\lim _{h \rightarrow 0} \frac{h}{e^{1 / h}+1}=0$.
$f(0-)=\lim _{h \rightarrow 0} \frac{-h}{e^{-1 / h}+1}=\lim _{h \rightarrow 0} \frac{-h}{1+\frac{1}{e^{1 / h}}}=0$
$f(0+)=\lim _{h \rightarrow 0} \frac{h}{e^{1 / h}+1}=0$.
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