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To approach this problem, I was thinking of doing l'hopital rule, but I don't think it works?

you have $\dfrac{e^{1/x}}x$. When you take derivative you're going to get

$$\dfrac{-x^{-2}e^{1/x}}1 = \dfrac{-e^{1/x}}{x^2}.$$

If i keep taking l'hopital rule of this problem, I will keep getting an $x^2$.

Is there another way? I'm sure the answer is right in front of me, I just can't see it.

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$t=1/x$ tends to $+\infty$ or $-\infty$ when $x\to 0^+$ or $x\to 0^-$ – Philippe Malot May 7 '13 at 20:58
I know, I keep getting an indeterminate form – Billy Thompson May 7 '13 at 20:59
$te^t$ doesn't give an indeterminate form when $t\to +\infty$. The other one is well known too. – Philippe Malot May 7 '13 at 21:02
e^1/x is going to 0, so wouldn't you have 0/0? – Billy Thompson May 7 '13 at 21:05
When $x \rightarrow 0$, the fraction $1/x$ is approaching infinity, not zero. That means $e^{1/x}$ is approaching infinity, as well. – Austin Mohr May 7 '13 at 21:07

If you make the change of variable $x\to z^{-1}$, as $x\to 0^-,z\to-\infty$ and in this case your limit turns into $\lim_{z\to-\infty}ze^z=\lim_{z\to-\infty}\dfrac{z}{e^{-z}}=0$ :)

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Let $t=-\frac{1}{x}$:

$$\lim_{x\to 0^-}\frac{e^{\frac{1}{x}}}{x}=\lim_{t\to\infty}\frac{-t}{e^t}=0$$

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How come you can change the the limit from 0- to infinity? – Billy Thompson May 7 '13 at 21:20
@BillyThompson: Because if $x$ approaches $0$ from the left, then $t=-\frac{1}{x}$ approaches infinity. – Jared May 7 '13 at 21:21

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