# finding the derivative

today I was thinking on a problem which states:

let $e(x)$ be a function defined by

$e(x)=e^{-\frac{1}{x}}$ if $x>0$ and $e(x)=0$ if $x\le 0$. the problem asks to prove the function is smooth. I tried to prove $\lim_{x\to 0^+}\frac{e^{-\frac{1}{x}}}{x}=0$ using the L'Hospitals rule, but applying that results in more $x$'s in the denominator. what should I do?

-

$$\displaystyle \lim_{x\to 0^+}\frac{e^{-\frac{1}{x}}}{x}=\displaystyle \lim_{x\to 0^+} \frac {\frac{1}{x}}{e^{\frac{1}{x}}}$$

Now substitute : $t= \frac {1}{x}$ , so:

$$\displaystyle \lim_{x\to 0^+} \frac {\frac{1}{x}}{e^{\frac{1}{x}}} =\displaystyle \lim_{t\to + \infty} \frac{t}{e^t}$$

Now apply L'Hopital rule , so :

$$\displaystyle \lim_{t\to + \infty} \frac{t}{e^t} = \displaystyle \lim_{t\to + \infty} \frac{1}{e^t}= 0$$

-
thanks, it was nice. –  Goodarz Mehr Feb 13 '12 at 10:45