# $0/0$ type limit question.

$$\lim_{x\to(0)} \frac{e^{1/x}}{x^2} =?$$

I used L'hopital but didn't solve.

• This is not $\frac{0}{0}$ indeterminate form; this is $\frac{\infty}{0}$ which essentially acts like $\infty \cdot \infty$. That realization alone should tell you that the limit will diverge to infinity. Dec 1, 2014 at 23:41
• @graydad I made a graph of it. It won't go to infinity if zero is approached from the left side. In fact, in that case, it indeed is a zero/zero form... Dec 1, 2014 at 23:55
• Good point; my statement is true for the right hand limit. However your graph is more evidence that the limit doesn't exist, as the right and left limits must match. Dec 2, 2014 at 0:02
• Actually i was forget to add "-". For this reason i said its 0/0 type. thanks anyway. Dec 2, 2014 at 18:56

This limit does not exist, because $1/x$ is discontinuous at $0$, and the numerator is either $+\infty$ or $0$.
$\lim_{x\to0^+}$ is infinite.
$\lim_{x\to0^-}$ is zero (it is the inverse of $\lim_{x\to0^+}$).
This limit doesn't exist. From the left the limit approaches $0$ and from the right it diverges to $+\infty$. I will now prove this.
Let's take the left : $$L^-=\lim_{x\to0^-}\frac{e^{1/x}}{x^2}$$ Let $y = 1/x$. Because as $y\to-\infty$, $1/y\to0^-$, $$L^-=\lim_{y\to-\infty}\frac{e^y}{1/y^2}=\lim_{y\to-\infty}\frac{y^2}{e^{-y}}=\lim_{y\to-\infty}\frac{2y}{-e^{-y}}=\lim_{y\to-\infty}\frac{2}{e^{-y}}=\lim_{y\to-\infty}2e^y$$ Clearly this approaches $0$ by the basic properties of exponents. Similarly it can be shown $$L^+ = \lim_{x\to0^=}\frac{e^{1/x}}{x^2} = \lim_{y\to+\infty}\frac{e^y}{1/y^2} = \lim_{y\to+\infty}2e^y$$ which diverges to $+\infty$. As $L^+ \neq L^-1$m the limit does not exist.