# Limit of $1/x^2$ - Apostol 3.2, Example 4

In Apostol, One Variable Calculus Volume 1, section 3.2, page 130, he gives the following example (roughly paraphrased):

Let $f(x) = \frac{1}{x^2}$ if $x \neq 0$, and let $f(0) = 0$. To prove rigorously that there is no real number $A$ such that $\lim_{x\to0^+} f(x)=A$, we may argue as follows: Suppose there were such an $A$, say $A>0$. Choose a neighborhood $N(A)$ of length $1$.

In the interval $0 < x < \frac{1}{A + 2}$, we have $f(x) = \frac{1}{x^2} > (A + 2)^2 > (A + 2)$, so $f(x)$ cannot lie in the neighborhood $N(A)$. Thus, every neighborhood $N(0)$ contains points $x > 0$ for which $f(x)$ is outside $N(A)$, so (3.3) is violated for this choice of $N(A)$. Hence $f$ has no right-hand limit at $0$.

While I intuitively understand why the function has no limit, I'm completely lost on his proof. What justifies him from moving from $f(x)$ lies outside of $N(A)$ for the neighborhood $0 < x < \frac{1}{A+2}$ to $f(x)$ lies outside every neighborhood?

The way I've been thinking about it is to translate the proof into terms from the $ε-δ$ definition. Thus, when he says "in the interval $0 < x < \frac{1}{A+2}$", he's setting $ε = \frac{1}{A+2}$, and then showing that $f(x)$ lies outside of $|f(x) - A| < ε$ for $|x-0| < δ$. But, if we were to prove that there is no limit, we have to show that for some $ε$, no $δ$ works. He says this in, "Thus, every neighborhood $N(0)$ contains points...", but I don't see how he can move from "this $δ$ doesn't work" to "no $δ$ works".

The best I can come up with is that either I'm making a mistake in thinking he's choosing $δ = \frac{1}{A+2}$, or that particular $δ$ is supposed to be a catch all. That is, if any $δ$ will work, this one should. But if that is the case, I don't see why this $δ$ has to be the one that works.

Thanks in advance.

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I've added LaTeX formatting to your question; take a look at the source to see how it works. In general, if you see a piece of LaTeX you want to know the code for on the site, you can right click on it and choose "Show Source" - this is a good way of picking up how to do things. – Zev Chonoles♦ Nov 29 '11 at 10:19
By the way, I would also like to applaud your in-depth explanation of your thoughts about your question - it is something that is unfortunately too rare. – Zev Chonoles♦ Nov 29 '11 at 10:21
Thanks for the formatting. I briefly tried to get it into LaTex, but I gave up after nothing worked.  Haha. I've tried to ask questions where I didn't explain as much, and the person I was asking either refused to help or started talking about a part of the problem that I wasn't asking about. In this case, I had already burned through all of my mathy friends, but they are all at least two years out from real analysis, and couldn't help very much. They did help me get this far, though. – Nathan Nov 29 '11 at 11:06

## 2 Answers

He shows for any $x$ with $0<x<{1\over A+2}$ that $f(x)>A+2$.

Now, for any neighborhood $N(0)$ you can select an $x\in N(0)$ with $0<x<{1\over A+2}$ ($N(0)$ has some radius $\delta$. Just choose some positive $x$ in the nhood that is less than ${1\over A+2}$). Then, from the above, $f(x)>A+2$ .

This is saying the same thing as "for any $\delta>0$ there is an $x$ with $|x-0|<\delta$ with $f(x)>A+2$".

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Oh! I think I get it. Any neighborhood N(0) will contain points such that 0<x<{1\over A+2}, and for those points f(x) > A+2. You can't get around that, no matter what δ you choose.   What makes me sure that I get it is that now I don't understand why I didn't see that in the first place. – Nathan Nov 29 '11 at 11:15
Great. And, I agree with the others. Thank you for a nicely thought out post. – David Mitra Nov 29 '11 at 11:16

First of all, I join Zev in appreciating your work.

The best I can come up with is that either I'm making a mistake in thinking he's choosing $\delta = \frac{1}{A+2}$, or that particular $\delta$ is supposed to be a catch all. That is, if any $\delta$ will work, this one should. But if that is the case, I don't see why this $\delta$ has to be the one that works.

Your second guess is almost correct. Let's see what exactly is going on.

Suppose some $\delta > 0$ works. That means that for all $x$ such that $0 < |x| < \delta$, $f(x)$ lies in the $1$-neighborhood of $A$. Now pick any $x$ such that $0 < |x| < \delta$ and $0 < x < \frac{1}{A+2}$ are both satisfied; this is equivalent to saying $$0 < x < \min \left\{ \delta, \frac{1}{A+2} \right\}.$$ Certainly there is at least one such $x$ (in fact, there are infinitely many $x$'s possible). For this $x$,

• since $0 < x < \frac{1}{A+2}$, we already know that $f(x)$ does not lie in the $1$-neighborhood of $A$.

• since $0 < |x| < \delta$ also holds, by our assumption, $f(x)$ lies in the $1$-neighborhood of $A$.

Obviously, these two conclusions contradict each other, which implies that our starting assumption must be wrong. That is why, no $\delta > 0$ can work.

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I think I get it.  Is this like saying that for a given δ1 such that for 0 < x < δ1, if δ1 does not work, than no δ > δ1 will work? – Nathan Nov 29 '11 at 11:03
@Nathan Your observation is accurate. But I am not sure that is very related to the answer to your question. – Srivatsan Nov 29 '11 at 11:06
On reflection, I'm not sure either. This is helping, though. Thanks! – Nathan Nov 29 '11 at 11:11
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