The function $f$ is given by
$$ f(x) = \begin{cases} e^{−\frac{1}{|x|}} & \text{if } x \ne 0\\ p & \text{if } x = 0 \end{cases}. $$
I have to find out which value of $p$ I need, so that the function is differentiable.
Left limit:
so what is the limit $$\lim_{x\to 0^{-}} -\frac{e^{-\frac{1}{x}}}{x^2}.$$
i know $e^{-1/x}$ limit is $0$ so is the total limit then also $0$?
Probably not because $-\frac1{x^2}$ is also there but how do I do it then?
And can somebody explain me how to make the formulas and stuff look pretty for next times?
A differentiable function must be continuous. Since $\lim_{x\to0}e^{-1/|x|}=0$, we get that $p=0$.
But is the function differentiable? Well, with $t=1/x$, $$ \lim_{x\to0^+}\frac{e^{-1/|x|}}{x}=\lim_{t\to\infty}{t}{e^{-t}}=0 $$ and with $t=-1/x$, $$ \lim_{x\to0^-}\frac{e^{-1/|x|}}{x}=\lim_{t\to\infty}{-t}{e^{-t}}=0 $$