Evaluating limit $lim_{xrightarrowinfty}2x(a+x(e^{-a/x}-1))$

I'm stumped by the following limit: $\lim_{x\rightarrow\infty}2x(a+x(e^{-a/x}-1))$

Mathematica gives the answer as $a^2$ but I'd like to know the evaluation steps. I've staring at it for a while, but can't figure it out. Seems like you can't use L'Hopital's rule on this one. I've tried substitution $y=1/x$ but that didn't help. I am probably not seeing something simple. Any hints?

Answer

HINT:

$$
\lim\limits_{x\to\infty} 2x \left(a + x \left(\mathrm{e}^{-a/x}-1\right)\right) = 2 a^2 \lim\limits_{x\to\infty} \frac{ \exp\left(-\frac{a}{x}\right) - 1 + \frac{a}{x} }{\left(\frac{a}{x}\right)^2} = 2 a^2 \lim\limits_{y\to 0^+} \frac{ \exp\left(-y\right) - 1 + y }{y^2}
$$
Now use l'Hospital's rule twice.