Oblique asymptote of ${1 \over 1-e^{-1 \over x}}$ $${1 \over 1-e^{-1 \over x}}$$
I know that m=$\lim_{x \to +\infty}$ $f(x)/x$ 
$\lim_{x \to +\infty}$ ${1 \over x-xe^{-1 \over x}}$
$e^{-1 \over+\infty}$ is $0^-$ ; then $1 \over x-x$ is 1; so m=1
But I learnt that when i substitute the x with $\infty$ I should do that for all the x, so I'd have at the denominator $\infty$-$\infty$, no good.
If i apply hopital i'll have 1/$\infty$ which is 0 and is wrong.
And q=$\lim_{x \to \infty}$ $f(x)-mx$, but I can't even solve the one above, so i can't do this.
Thanks.

Hints:
$$\begin{align}&(1)\;\;\;\frac{f(x)}x=\frac1{x(1-e^{-1/x})}=\frac{e^{1/x}}{x(e^{1/x}-1)}\\{}\\&(2)\;\;\;\text{Substitute}\;\;h:=\frac1x\implies\frac1{x(e^{1/x}-1)}=\frac h{e^h-1}\\{}\\
&(3)\;\;\;\text{What is}\;\;\left(e^0\right)'\;?\end{align}$$
