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.
 A: 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. 
A: 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 $f(x)>A+2$".
A: I've just spent a couple of hours banging my head against the same proof. The key is that the neighbourhood of 1 is not particularly important, nor is the subdomain $0 < x < \frac{1}{A+2}$.
To demonstrate, consider the subdomain
$$0 < x < \frac{1}{A+1}$$
$$u = \frac{1}{A+1} \Rightarrow \frac{1}{u^{2}} = (A + 1)^{2} \Rightarrow \frac{1}{x^{2}} > (A + 1)^{2}$$
Since $A \geq 0$
$$(A + 1)^{2} \geq (A + 1)$$
Hence
$$\frac{1}{x^{2}} > A + 1$$
And
$$| \frac{1}{x^{2}} - A | > 1$$
Thus when $\epsilon$, which can be chosen to be arbitrarily small, is chosen to be less than or equal to 1, the epsilon expression is violated throughout the subdomain under consideration. Since the delta expression necessarily intersects with this subdomain, regardless of the value of delta, the limit does not exist.
In fact any $\frac{1}{A + k}; k \geq 1$ should work equally well, since the key step is the simplifying inequality from the squared to the non squared form. The neighbourhood of size 1 (i.e. $\epsilon = \frac{1}{2}$) seems to be more or less spurious.
