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?
Hint: It may be fruitful to substitute $\alpha = 1/x$, in which case you obtain the limit

$$ \lim_{ \alpha \rightarrow \infty} \left(2 - \frac{1}{\alpha} \right) \alpha^3 e^{\alpha - \alpha^2} $$

I should note that, here, I'm taking your limit to in fact be the limit as $x$ approaches $0$ from the positive direction. If you're intending for your limit to be two-sided, then you should think about why that would cause problems.
$$\lim_{x\to0}(f(x)g(x)) = \lim_{x\to0}(f(x)) \cdot \lim_{x\to0}(g(x))  $$

With that being said you can let $f(x) =  (2-x)/x^3 $ and $ g(x) = e^{(x-1)/x^2}  $

I hope this helps.