Checking differentiability of $e^\frac{-1}{x}$ 
Let $f:\mathbb R\to\mathbb R$ is defined by
  $$f(x)=
\begin{cases}
e^\frac{-1}{x}&&x>0\\
0&&x\le0
\end{cases}$$

How do I check differentiability of $f(x)$ at $x=0$?
I have tried to use first principle but cannot proceed.
 A: First principles:
From the perhaps most important inequality for the exponential,  $e^t\ge 1+t$ for all $t\in\Bbb R$, we find $e^t=(e^{t/2})^2\ge (1+\frac12t)^2=1+t+\frac14t^2$ for $t\ge 0$. 
Thus for $x>0$,
$$\frac{f(x)-f(0)}x=\frac 1xe^{-\frac1x}=\frac1{xe^{1/x}}\le\frac1{x(1+\frac1x+\frac1{4x^2})}=\frac{x}{x^2+x+\frac1{4}}\to 0$$
A: You apply repeatedly the fact that if $f$ is defined in a nbd of $x_0$ and the derivative $f'(x)$ has a limit as $x\to x_0$ then $f$ is differentiable at $x_0$, because by Darboux's theorem, the derivative function cannot have removable discontinuities.
A: Use L'Hôpital's rule repeatedly as you approach $x = 0$ from the right:
\begin{align*}
\lim\limits_{x \to 0^+}f'(x) &= \lim\limits_{x \to 0^+}\frac{e^{-\frac{1}{x}}}{x^2} \\
&= \lim\limits_{x \to 0^+}\frac{\frac{1}{x^2}}{e^{\frac{1}{x}}} \quad\text{limit of the form $\frac{\infty}{\infty}$} \\
&= \lim\limits_{x \to 0^+}\frac{-\frac{2}{x^3}}{e^{\frac{1}{x}} \cdot \big(-\frac{1}{x^2}\big)} \\
&= \lim\limits_{x \to 0^+}\frac{\frac{2}{x}}{e^{\frac{1}{x}}} \quad\text{limit of the form $\frac{\infty}{\infty}$} \\
&= \lim\limits_{x \to 0^+}\frac{-\frac{2}{x^2}}{e^{\frac{1}{x}} \cdot \big(-\frac{1}{x^2}\big)} \\
&= \lim\limits_{x \to 0^+}\frac{2}{e^{\frac{1}{x}}} \\
&= \lim\limits_{x \to 0^+}2e^{-\frac{1}{x}} = 0 \\
\end{align*} Of course, approaching $x = 0$ from the left, $f(x)$ is just the constant function $0$; so $$\lim\limits_{x \to 0^-}f'(x) = \lim\limits_{x \to 0^-} 0 = 0$$ Thus, $f(x)$ is differentiable at $x = 0$ and the value of its derivative there is $0$.
A: This function is actually indefinitely derivable on $\mathbb{R}$.
By induction, you can show the k-th derivative exists on $\mathbb{R}^*$ and is given


*

*on $(-\infty, 0)$ by $x\mapsto 0$,

*on $(0,+\infty)$ by a function $x \mapsto P(1/x)e^{-1/x}$, for some polynomial $P\in \mathbb{R}[t]$.


Then by induction, you can show the $k$-th derivative at $0$ is $0$, using that for any polynomial $P\in \mathbb{R}[t]$,$P(t)/exp(t)$ tends to $0$ as $t\to +\infty$.
