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

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.

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 $f(x)>A+2$".