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

I used L'hopital but didn't solve.

Huseyin
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    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. – graydad Dec 01 '14 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... – imranfat Dec 01 '14 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. – graydad Dec 02 '14 at 00:02
  • Actually i was forget to add "-". For this reason i said its 0/0 type. thanks anyway. – Huseyin Dec 02 '14 at 18:56

2 Answers2

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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 $\lim_{x\to0^+}$).

2

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.

David Etler
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