derivability of $f$ at $x=0$ Let
$$f(x) = \left\{\begin{array}{ll}
e^{\frac{x-1}{x^2}} & \text{if } x\neq 0\\
%\\
0 & \text{if } x=0
\end{array}\right.
$$
I want to study the derivability of $f$ at $x=0$. By applying Hopital rule, we get
$$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\Big( \frac{2-x}{x^3}\Big)e^{\frac{x-1}{x^2}}.$$
I think
$$\lim_{x\to 0}\Big( \frac{2-x}{x^3}\Big)$$
does not exist
 A: Since $\,e^z\geqslant 1+z\;$ for all $\;z\in\mathbb R\,,\,$ it follows that
$e^{\frac t2}\geqslant 1+\dfrac t2>\dfrac t2\;$ for all $\;t\in\mathbb R\;,\;$ hence ,
$t<2e^{\frac t2}\;$ for all $\;t\in\mathbb R\;,$
$0<\dfrac t{e^t}<\dfrac{2e^{\frac t2}}{e^t}=\dfrac2{e^{\frac t2}}\;$ for all $\;t>0\,.\quad\color{blue}{(1)}$
Since $\,\lim_\limits{t\to+\infty}\dfrac2{e^{\frac t2}}=0\,,\,$ by applying Squeeze Theorem to the inequalities $\,(1)\,,\,$ we get that $\;\lim_\limits{t\to+\infty}\dfrac t{e^t}=0\,.$
Therefore ,
$\lim_\limits{x\to0}\dfrac{1-x}{x^2}e^{\frac{x-1}{x^2}}=\lim_\limits{x\to0}\dfrac{\frac{1-x}{x^2}}{e^{\frac{1-x}{x^2}}}\!\!\underset{\overbrace{\text{by letting }t=\frac{1-x}{x^2}}}{=}\lim_\limits{t\to+\infty}\dfrac t{e^t}=0\;,\;$ hence ,
$\begin{align}
\lim_\limits{x\to0}\dfrac{f(x)-f(0)}{x-0}&=\lim_\limits{x\to0}\dfrac1x e^{\frac{x-1}{x^2}}=\lim_\limits{x\to0}\dfrac x{1-x}\dfrac{1-x}{x^2}e^{\frac{x-1}{x^2}}=\\
&=\lim_\limits{x\to0}\dfrac x{1-x}\cdot\lim_\limits{x\to0}\dfrac{1-x}{x^2}e^{\frac{x-1}{x^2}}=0\cdot0=0\,.
\end{align}$
Consequently $\,f(x)\,$ is differentiable at $\,x=0\,$ and it results that $\,f’(0)=0\,.$
