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.