# Showing that the $\lim_{x\to 0}\frac{1}{x^2}$ does not exist

I want to show that the $\lim_{x\to 0}\frac{1}{x^2}$ does not exist.

I start with $|x-0|<\delta$ and $|\frac{1}{x^2}|<\epsilon$

I let $\epsilon = 1$ so $|\frac{1}{x^2}|<1$

Then $|1|<x^2$ which means $|1|<x$

If I make my $\delta= 1$

$|x-0|<1$

and $|x|<1$

I can't have $|x|<1$ and $|1|<x$ so I can't have a limit here. Is that right?

• The limit exists and is $+\infty$. Dec 17, 2015 at 14:40
• Thanks. I gathered that from my textbook. TO give a bit of context, the textbook (Stewart's calculus) is using this as a function that is not continous. It states that it is not continous because the limit does not exist at 0. So, I was wondering how I could show this. Dec 17, 2015 at 14:41
• @EmilioNovati It is quite well possible that the OP is working in $\mathbb R$. In that context the limit does not exist. Dec 17, 2015 at 14:42
• I want to know why that is? Can it be shown in a straightforward way? Dec 17, 2015 at 14:42
• Line number 2 assumes that $\delta$ is a very small positive number to indicate that $x \tends 0$. In that case, $\abs \frac{1}{x^2} < \epsilon$ has to be a very very large number and cannot be "let equal to 1" as you have done in your third line. That is where you have gone wrong. Dec 17, 2015 at 14:54

Suppose that the limit exists and equals $c\in\mathbb{R}$.

Then for e.g. $\epsilon>1$ some $\delta>0$ must exist with $\left|x\right|<\delta\implies\left|\frac{1}{x^{2}}-c\right|<1$.

However, if we take $\left|x\right|$ small enough then $\left|\frac{1}{x^{2}}-c\right|$ will definitely exceed $1$ (do you see why?).

We conclude that the limit does not exist.

• Fantastic. Yes, I see why, as x goes below 1, the numerator starts to increase and will increase indefinitely as x decreases. Dec 17, 2015 at 15:01
• Yes, you are correct. Glad to help. Dec 17, 2015 at 15:04

Actually the limit exists and is equal to $\infty$. No matter from which direction you approach $0$ (from negative values or from positive values) the term $x^2$ is positive. Moreover as $x \to 0$, the term $x^2$ becomes very small, so that $\frac1{x^2}$ becomes very big (in other words, it grows steadily to infinity). Bringing all these together $$\lim_{x\to 0-}\frac1{x^2}=\lim_{x\to 0+}\frac{1}{x^2}=+\infty$$

• Thanks Stef, I just realized that I did not pay attention to the "square" of the denominator and have promptly removed my mistaken answer. Dec 17, 2015 at 14:57
• $\pm\infty$ are often not considered to be "real limits" as in "the limit is a real number". One might say that for $x\to 0$ the function is strictly divergent with $\lim\limits_{x\to 0} \frac{1}{x^2}=\infty$. Dec 17, 2015 at 14:57
• @Hirshy Yes, I agree with you. I just wanted to distinguish between "it doesn't exist" and "it goes to infinity". Thanks for correcting it. Dec 17, 2015 at 14:58

By definition, the limit of a function at a point is defined as

$$\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \to \left| {f(x) - L} \right| < \varepsilon } \right)} \right)$$

and its negation will be (see this post)

$$\forall L,\exists \varepsilon > 0:\left( \forall \delta > 0,\exists x:(0 < \left| x - a \right| < \delta \wedge |f(x) - L| \ge \varepsilon \right))$$

So for your special case we shall prove that

$$\forall L,\exists \varepsilon > 0: ( \forall \delta > 0,\exists x:(0 < \left| x \right| < \delta \wedge |\frac{1}{x^2} - L| \ge \varepsilon )$$

You can see this as a game! Your opponent chooses $$L$$ then you select an $$\epsilon$$. Next, the opponent chooses $$\delta$$ and then you choose the $$x$$. You should be wise enough to make proper choices so that you will win the game no matter what your opponent chooses. So your choices must satisfy the conditions $$\left| x \right| < \delta \wedge |\frac{1}{x^2} - L| \ge \varepsilon$$ regardless of opponent's choices. So, let us play!

The opponent makes a choice for $$L$$. we choose $$\varepsilon$$ to be $$1$$. Then the opponent selects a $$\delta$$ and we should make our final choice for $$x$$ such that

$$\left| x \right| < \delta \wedge |\frac{1}{x^2} - L| \ge 1$$

we can choose $$x_n=\frac{1}{\sqrt{n+L}}$$ and our conditions turns to be

$$\left| \frac{1}{\sqrt{n+L}} \right| < \delta \wedge |n| \ge 1$$

So, no matter what the opponent chooses, we will select large enough values for $$n$$ such that the above conditions are satisfied and we win the game! Hence, the limit does not exist!

Find the following limit: $$\lim _{x \rightarrow 0} \frac{1}{x^{2}}$$ $$\lim _{x \rightarrow 0} \frac{1}{x^{2}}=\lim _{x \rightarrow 0} \exp \left(\log \left(\frac{1}{x^{2}}\right)\right):$$ $$\lim _{x \rightarrow 0} \exp \left(\log \left(\frac{1}{x^{2}}\right)\right)$$ $$\exp \left(\log \left(\frac{1}{x^{2}}\right)\right)=\exp (-2 \log (x)):$$ $$\lim _{x \rightarrow 0} \exp (-2 \log (x))$$ $$\lim _{x \rightarrow 0} \exp (-2 \log (x))=\exp \left(\lim _{x \rightarrow 0}-2 \log (x)\right):$$ $$\exp \left(\lim _{x \rightarrow 0}-2 \log (x)\right)$$ Applying the product rule, write $$\lim _{x \rightarrow 0}-2 \log (x)$$ as $$-2 \lim _{x \rightarrow 0} \log (x)$$ : $$\exp \left(-2 \lim _{x \rightarrow 0} \log (x)\right)$$ $$\lim _{x \rightarrow 0} \log (x)=-\infty:$$ $$e^{-2[-\infty}$$ $$e^{-2(-\infty)}=\infty:$$ $$\lim_{x\to 0}\frac{1}{x^2}: \infty$$