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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?

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hint: expand $e^{-a/x}$ to second order – Raymond Manzoni Aug 29 '12 at 16:15
Aha!!! I see it now. Thank you, Raymond! – M.B.M. Aug 29 '12 at 16:19
You are welcome! – Raymond Manzoni Aug 29 '12 at 16:26
up vote 3 down vote accepted


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

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Hint $\, $ For $\rm\, z = \displaystyle-\frac{a}x\: $ it's $\displaystyle\rm\:2a^2 \lim_{z\to 0^{-}}\frac{f(z)-(f(0)+f'(0)\,z)}{z^2}\, =\, a^2\, f''(0),\ \ f = e^z,\, $ by Taylor / L'Hospital.

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