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What is $$\lim_{x\to0}\frac0{x^2}$$

I would guess that it is $0$, because $0$ divided by anything is $0$.

$x$ is just a very very small number like $0.000001$. And in the case of $\cfrac1{x^2}$ the limit should be $\infty$ right?

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    $\begingroup$ Yes. It might help to plot the graphs of these functions to see things more clearly. $\endgroup$ – Dirk May 10 '17 at 9:58
  • $\begingroup$ uses l'hopital $\endgroup$ – Saketh Malyala May 10 '17 at 10:17
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Yes, you are right.

Your function is constant and it is defined on $\mathbb R\setminus\{0\}$, so it goes to $0$ when $x$ goes to $0$.

And in the case $f(x)=1/x^2$, the limit is indeed $+\infty$.

Look at the graph to convince yourself.

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but $$\frac0{x^2}=0$$ if $$x\ne0$$ thus the Limit is 0

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The limit is indeed $0$, as

$$\forall\,\epsilon>0:\exists\,\delta>0:\forall\,x:0<|x-0|<\delta\implies\left|\frac0{x^2}-0\right|<\epsilon.$$

The last inequality always holds and any $\delta$ can do.

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