How can I calculate this limit? ( I tried l'Hopital and failed ) I have to calculate this : $$ \lim_{x\to 0}\frac{2-x}{x^3}e^{(x-1)/x^2} $$ Can somebody help me?

Letting $w=1/x$, we have
$$
\lim_{x\downarrow 0}\frac{2-x}{x^3}e^{(x-1)/x^2} = \lim_{w\to+\infty} \left(2 - \frac 1 w \right) w^3 e^{w^2\left(\frac 1 w - 1\right)} = \lim_{w\to+\infty} (2w^3 - w^2) e^{w-w^2}
$$
$$
= \lim_{w\to+\infty} \frac{2w^3-w^2}{e^{w^2-w}}.
$$
L'Hopital should handle that.
Maybe I'll post something on $x\uparrow 0$ later . . .
