prove that $ \lim_{x \to 0} \frac{e^{1/x}}{x}$ does not exist. By Substitution of y = $\frac {1}{x}$ i have managed to show that
$\lim_{x \to 0^+}\frac{e^{1/x}}{x} = \infty $
but i can't find a way to show that 
$\lim_{x \to 0^-}\frac{e^{1/x}}{x} = 0 $
I've tried L'Hospital rule but ended with
$\lim_{x \to 0^-}-\frac{e^{1/x}}{x^2}$ = -"$\frac {0}{0}$" again.
I found by deriving the function that in a small left environment of 0 the function is monotonously increasing.
 A: HINT: 
$$\lim_{x \to 0^-}\frac{e^{1/x}}{x}  = \lim_{x \to 0^+}-\frac{e^{-1/x}}{x} = \lim_{y \to +\infty}-\frac{e^{-y}}{1/y} =  \lim_{y \to +\infty}-\frac{y}{e^{y}}.$$
A: The substitution $yx=1$ will work fine in both cases. In one, you should get $y\to+\infty$ and in the other $y\to-\infty$. In both cases you'll have to look at $\psi(y)=ye^{y}$
A: $\displaystyle e^{\frac{1}{x}}=\sum_{i=0}^{\infty}\frac{1}{i!}\frac{1}{x^i}\Rightarrow \frac{e^{1/x}}{x}=\frac{1}{x}\sum_{i=0}^{\infty}\frac{1}{i!}\frac{1}{x^i}= \sum_{i=0}^{\infty}\frac{1}{i!}\frac{1}{x^{i+1}}$
Now you can easily show that $\displaystyle \sum_{i=0}^{\infty}\frac{1}{i!}\frac{1}{x^{i+1}}$ is unbounded.So the limit does not exist.
A: Using $e^t\ge 1+t$,
$$ 0\ge\lim_{x\to 0^-}\frac{e^{1/x}}{x}=\lim_{x\to-\infty}xe^{x}=\lim_{x\to+\infty}-\frac x{e^x}=\lim_{x\to+\infty}-\frac x{e^{x/2}e^{x/2}}\ge \lim_{x\to+\infty}-\frac{x}{(1+x/2)(1+x/2)}=0.$$
A: Using the inequality $e^x >1+x, \ x \in \mathbb{R}$, $\lim_{x \to 0^{+}}\frac{1}{x}+\frac{1}{x^2}=+\infty, \ \lim_{x \to 0^{-}}\frac{1}{x}-\frac{1}{x^2}=-\infty$ 
