# 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.

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 $$x>0$$, $$\frac{f(x)-f(0)}x=\frac 1xe^{-\frac1x}=\frac1{xe^{1/x}}\le\frac1{x(1+\frac1x+\frac1{4x^2})}=\frac{x}{x^2+x+\frac1{4}}\to 0$$
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.
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 $$0$$.
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 $$t\to +\infty$$.