Show that $\lim_{x \to 0} \frac{1}{x^2}$ does not exist.

I have seen this: Epsilon-delta proof that $ \lim_{x\to 0} {1\over x^2}$ does not exist but there is a step in the answer that I am not fully grasping. I wanted to try my own argument and see if I could convince myself.

In my argument, I use a step I am not sure I am allowed to use. If my method is simply not going to work, please let me know and I will modify the question to ask for clarification on the other answer. Also (if possible) I would prefer a hint than an answer unless I end up completely stuck!

I want to show $\exists \epsilon > 0 \forall \delta > 0$ s.t. whenever $|x| < \delta$, $|\frac{1}{x^2} - L| \geq \epsilon$

Side work: $x < \delta$

$x^2 < \delta^2$

$-x^2 > -\delta^2$ and $\frac{1}{\delta^2} < \frac{1}{x^2}$

$-Lx^2 > -L \delta^2$

$1 - Lx^2 > 1 - L \delta^2$

With this being done, choose $\epsilon = \frac{1 - L \delta^2}{\delta^2}$. This is the part I am unsure of. I have to show this is true $\forall \epsilon > 0$, and because there is a $\delta$ in both the numerator and denominator, I'm not sure this choice of $\epsilon$ is arbitrarily small.

If I continue I would have

$| \frac{1}{x^2} - \frac{Lx^2}{x^2}| = |\frac{1 - L^2x^2}{x^2}| > |\frac{1 - L^2\delta^2}{\delta^2}| = \epsilon$

The only issue is whether or not I can use that $\epsilon$. I am leaning towards no because if I break this back up into two fractions:

$|\frac{1}{\delta^2} - \frac{L\delta^2}{\delta^2}| \geq \frac{1}{\delta^2} - L$, and the first fraction becomes arbitrarily large due to $\delta$ being arbitrarily small in the denominator.

Is there a way I can modify my method to get it to work? Or is this doomed to fail?

Thanks in advance!

  • $\begingroup$ Given any epsilon, there exists at least one delta such that....... You wrote wrong at the begining. $\endgroup$
    – Bob Dobbs
    Commented Jul 10, 2022 at 20:52
  • 1
    $\begingroup$ @MehmetKırdar I'm not sure I follow. I want to show discontinuous. Therefore, I invert all of the quantifiers. That would change the $\exists$ to a $\forall$. I did a double take and googled this to confirm. $\endgroup$
    – Nolan P
    Commented Jul 10, 2022 at 21:24
  • $\begingroup$ If you are trying to prove that epsilon has some property for ALL $\delta$ you can't define epsilon in terms of any one specific $\delta$. $\endgroup$
    – fleablood
    Commented Jul 11, 2022 at 1:28

1 Answer 1


Well, you know that the limit is $\infty$, so it cannot be finite.

Suppose the limit exists (finite) and is $l$. By permanence of sign you know that $l\ge0$.

Let $\varepsilon>0$. Then, by assumption, there exists $\delta>0$ such that, for $0<|x|<\delta$, it holds $$ \left|\frac{1}{x^2}-l\right|<\varepsilon $$ The inequality is the same as $$ |1-lx^2|<\varepsilon x^2 $$ hence $$ -\varepsilon x^2<1-lx^2<\varepsilon x^2 $$ In particular, $(l+\varepsilon)x^2>1$, which is the same as $$ |x|>\sqrt{\frac{1}{l+\varepsilon}} $$ which is a contradiction, because it doesn't hold for $$ x=\frac{1}{2}\min\left(\delta,\sqrt{\frac{1}{l+\varepsilon}}\,\right) $$

You're simply starting wrong: you don't want to find some $\varepsilon>0$; you want to see that it's contradictory to assume that, for every $\varepsilon>0$, …

In a slightly different way. If $$ \lim_{x\to0}f(x)=l $$ then there exists $\delta>0$ such that $f$ is bounded over $(-\delta,\delta)\setminus\{0\}$. But $f(x)=1/x^2$ is unbounded over any punctured neighborhood of $0$.

  • $\begingroup$ I have one thing that I want to clarify. One, to make sure I understand your proof, at the end, by choosing x to be the minimum out of those things, you ensure that one of them will not hold. One of our choices is too small, and x will be larger. Is that correct? Also, I completely see where I went wrong. I forgot that when the power is even, $\frac{1}{x^n}$ converges to infinity. I was thinking about the odd powers where the limit does not exist, which is why I went to the converse. I actually think I could now prove this for $\frac{1}{x^{2n}}$ now. Thank you! $\endgroup$
    – Nolan P
    Commented Jul 10, 2022 at 21:30
  • $\begingroup$ @NolanP By assumption, the inequality $|x^{-2}-l|<\varepsilon$ holds for every $x$ such that $0<x<\delta$, but it can't hold for that particular number, which lies in $(0,\delta)$ and this is the sought contradiction. $\endgroup$
    – egreg
    Commented Jul 10, 2022 at 21:32
  • $\begingroup$ I understand the argument, it makes sense, what I'm wondering is why it doesn't hold for that particular number? It's not immediately obvious to me why that's the case. If it does hold the argument makes sense, I'm just wondering why it's clear that that number makes it fail. $\endgroup$
    – Nolan P
    Commented Jul 10, 2022 at 21:38
  • $\begingroup$ @NolanP The number is positive and less than $\delta$, so $|x^{-2}-l|<\varepsilon$ holds by assumption, but it's also less than $\sqrt{1/(l+\varepsilon)}$, so $|x^{-2}-l|<\varepsilon$ doesn't hold (by calculation). $\endgroup$
    – egreg
    Commented Jul 10, 2022 at 21:43
  • $\begingroup$ Oh I see it! $|\frac{1}{(\sqrt{1/2(l+\epsilon)})^2} - l| = |\frac{1}{1/4(l+\epsilon)} - l| = 4l + 4\epsilon - l| = |3l + 4\epsilon| > \epsilon$. That is a contradiction. Thank you! $\endgroup$
    – Nolan P
    Commented Jul 10, 2022 at 21:49

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