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! 

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
$$-\frac{a}{x}-\log x=\frac1x\left(-a-x\log x\right)\to+\infty\cdot(-a-0)=-\infty$$
