# Showing that a function is differentiable at $0$ with $f'(0)=0$

The function is defined by

$$f(x) = \begin{cases} e^{-1/x^{2}} & \text{if x \neq 0} \\ 0 & \text{if x = 0} \end{cases}$$

I tried using this definition

$$f'(a) = \lim_{x \to a} \frac {f(x)-f(a)}{x-a}$$

which results in having

$$\lim_{x \to 0} \frac {e^{-1/x^{2}}-0}{x-0} = \lim_{x \to 0} \frac {e^{-1/x^{2}}}{x}$$

I see that you can solve this using L'Hospital's rules, but the value is still $$\frac 00$$ no matter how many times I do the differential to each of the fraction.

What is the best way to approach this?

$$\lim_{x\to0}\frac{e^{-1/x^2}}x=\lim_{x\to\pm\infty}e^{-x^2}{x}=\lim_{x\to\pm\infty}\frac{x}{e^{x^2}}=\lim_{x\to\pm\infty}\frac{1}{2xe^{x^2}}$$ Explaination $$\lim_{x\to0}f(x)=\lim_{x\to\pm\infty}f(1/x)$$ It can be easily proved by using epsilon-delta language.
• I'm sorry this may sound embarrassing, even though I'm a 3rd year math major. Can you elaborate more on your hint? How does $\lim_{x\to0}\frac{e^{-1/x^2}}x$ becomes $\lim_{x\to\infty}e^{-x^2}{x}$? – Jeffry Santosa Dec 7 '18 at 6:12
• Note that $\lim_{x\to 0} f(x) = \lim_{x \to \infty} f(1/x)$ is not true in general. If the function is not continuous at zero the latter limit may still exist. – Pjotr5 Dec 7 '18 at 8:09
• @Pjotr5 Here $x\to\infty$ denotes $x\to\pm\infty$. – Kemono Chen Dec 7 '18 at 8:10
We have $$\lim_{x\rightarrow 0}\frac{e^{-1/x^2}}{x}=\lim_{t\rightarrow \infty}\frac{t}{e^{t^2}}=\lim_{t\rightarrow \infty}\frac{1}{2te^{t^2}}=0$$ by $$t=\frac{1}{x}$$ and L'Hospital's rules.