How to find $\lim_{x\rightarrow 0^+} \frac{e^{-a/x}}{x}$? I just review the following problem:
How to find the limits $\lim\limits_{h\rightarrow 0} \frac{e^{-h}}{-h}$ and $\lim\limits_{h\rightarrow 0} \frac{|\cos h-1|}{h}$?
However, I cannot still know how to solve the following: 
How to find the following: $$\lim_{x\rightarrow 0^+} \frac{e^{-a/x}}{x}, \ \ a>0$$
By L'hospital's rule:
$$\lim_{x\rightarrow 0^+} \frac{e^{-a/x} \frac{a}{x^2}}{1}= \lim_{x\rightarrow 0^+} \frac{e^{-a/x} a}{x^2}$$ 
it seems that the degree of the denominator will increase; however, I am still confused about the limit of this problem. Please advise, thanks! 
 A: Note that for $a>0$
$$ \frac{e^{-a/x}}{x}=e^{\log e^{-a/x}-\log x}=e^{-\frac{a}{x}-\log x}\to e^{-\infty}=0$$
indeed
$$-\frac{a}{x}-\log x=\frac1x\left(-a-x\log x\right)\to+\infty\cdot(-a-0)=-\infty$$
A: A possible approach without L'Hôpital's rule: Since
$$
 e^y = \sum_{n=0}^\infty \frac{y^n}{n!} > \frac{y^2}{2}
$$
for $y > 0$, we have
$$
0 < \frac{e^{-a/x}}{x} = \frac{1}{xe^{a/x}} < \frac{2x}{a^2}
$$
so that the limit is zero. The same argument can be used to show that
$$
 \lim_{x \to 0^+} \frac{e^{-a/x}}{x^k} = 0
$$
for $a > 0$ and any exponent $k > 0$. The essential idea is that
the exponential function "grows faster than any polynomial".
A: We can use the standard result $x^n/e^x\to 0$ as $x\to\infty$. For the current question just put $a/x=t$ so that $t\to\infty$ and the given expression equals $(1/a)(t/e^t)$ which tends to $(1/a)\cdot 0=0$.
A: \begin{align*}
\lim_{x\rightarrow 0^{+}}\dfrac{e^{-a/x}}{x}&=\lim_{x\rightarrow 0^{+}}\dfrac{x^{-1}}{e^{a/x}}\\
&=\lim_{x\rightarrow 0^{+}}\dfrac{-x^{-2}}{e^{a/x}\cdot-ax^{-2}}\\
&=\lim_{x\rightarrow 0^{+}}\dfrac{1}{a}\dfrac{1}{e^{a/x}}\\
&=\dfrac{1}{a}\cdot 0\\
&=0.
\end{align*}
Another way:
\begin{align*}
\lim_{x\rightarrow 0^{+}}\dfrac{e^{-a/x}}{x}&=\lim_{u\rightarrow\infty}ue^{-au}\\
&=\lim_{u\rightarrow\infty}\dfrac{u}{e^{au}}\\
&=\lim_{u\rightarrow\infty}\dfrac{1}{ae^{au}}\\
&=0.
\end{align*}
A: First note that the derivative approach isn't going to fly, $e^{-a/x}$ is not even continuous at $0$!
Rewrite your limit as
$$
\lim_{x\to 0^+}\frac{1/x}{e^{a/x}}
$$
note that both top and bottom tend to positive infinity. Apply L'Hopital's, 
$$
\lim_{x\to 0^+}\frac{1/x}{e^{a/x}}=
\lim_{x\to 0^+}\frac{-1/x^2}{-a/x^2e^{a/x}}\\
=\frac1 a\lim_{x\to 0^+}\frac{1}{e^{a/x}}=0
$$
