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

I used L'hopital but didn't solve.

  • 1
    $\begingroup$ 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. $\endgroup$
    – graydad
    Dec 1, 2014 at 23:41
  • $\begingroup$ @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... $\endgroup$
    – imranfat
    Dec 1, 2014 at 23:55
  • $\begingroup$ 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. $\endgroup$
    – graydad
    Dec 2, 2014 at 0:02
  • $\begingroup$ Actually i was forget to add "-". For this reason i said its 0/0 type. thanks anyway. $\endgroup$
    – Huseyin
    Dec 2, 2014 at 18:56

2 Answers 2


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.


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