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Show that the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $$f(x)=\begin{cases} 0,& \text{if } ~x=0 \\ \frac{x}{1+e^{1/x}}, & x\ne 0. \end{cases} $$ is continuous at $x=0$ but not differentiable at $x=0$.

Please help me.

$\lim_{x\rightarrow 0+}\frac{x}{1+e^{1/x}}=?$

$\lim_{x\rightarrow 0-}\frac{x}{1+e^{1/x}}=?$

$Rf'(0)=\lim_{x\rightarrow 0+}\frac{f(h)-f(0)}{h}=\lim_{x\rightarrow 0+}\frac{1}{1+e^{1/h}}=?$

$Lf'(0)=\lim_{x\rightarrow 0-}\frac{f(h)-f(0)}{h}=\lim_{x\rightarrow 0-}\frac{1}{1+e^{1/h}}=?$

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Hint: prove that $$ \lim_{h\to0^-}e^{1/h}=0, \qquad \lim_{h\to0^+}e^{1/h}=\infty $$

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    Yes, I know, but please help me find those. – 2017-02-20
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    @user1942348 Apply the well known $\lim_{t\to-\infty}e^t=0$, $\lim_{t\to\infty}e^t=\infty$. – 2017-02-20
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Continuity at $x=0$:

Since $e^{1/x}>0$ for all $x$ we have:

$0 \le |f(x)|=\frac{|x|}{1+e^{1/x}} \le |x|$ for all $x \ne 0$

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    @Ered ok, great. Hope you suggest sandwich theorem next. what next to find f'(x) – 2017-02-20