To show in general that $f$ approaches $1/a$ near $a$ for any $a$ we proceed in basically the same way, except that,again, we have to be a little more careful in formulating our initial stipulation. It's not good enough simply to require that $|x-a|$ should be less than $1$, or any other particular number, because if $a$ is close to $0$ this would allow values of $x$ that are negative (not to mention the embarrassing possibility that $x=0$, so that $f(x)$ isn't even defined!).
The trick in this case is to first require that
$$|x-a|<\frac{|a|}{2};$$
in other words, we require that $x$ be less than half as far from $a$ as $a$ is from $0$. You should be able to check first that $x \ne 0$ and that $1/|x|$ < $2/|a|$, and then work out the rest of the argument.
In order to check that $x \ne 0$ I checked both cases $a \ge 0$ and $a<0$:
$a > 0$:
$|x-a|<\dfrac{|a|}{2}$ $\implies$ $-\dfrac{|a|}{2}<x-a<\dfrac{|a|}{2}$
$\implies$ $-\dfrac{a}{2}<x-a<\dfrac{a}{2}$
$\implies$ $-\dfrac{a}{2}+a<x<\dfrac{a}{2}+a$ and so
$\dfrac{2a-a}{2}<x<\dfrac{a+2a}{2}$. Thus $x>0$ since $0<\dfrac{a}{2}<x$.
$a<0$:
$|x-a|<\dfrac{|a|}{2}$ $\implies$ $-\dfrac{(-a)}{2}<x-a<\dfrac{(-a)}{2}$
$\implies$ $-\dfrac{(-a)}{2} + a <x<\dfrac{(-a)}{2}+a$
$\implies$ $\dfrac{a}{2}+a=\dfrac{2a +a}{2}=\dfrac{3a}{2}<x<\dfrac{(-a)}{2}+a=\dfrac{2a-a}{2}=\dfrac{a}{2}$.
Therefore $x<0$ since $x<\dfrac{a}{2}<0$.
In order to get that $1/|x|<2/|a|$, I did the following:
$|x-a|<\dfrac{|a|}{2}$ $\implies$ $-\dfrac{|a|}{2}<x-a<\dfrac{|a|}{2}$
and so $-\dfrac{|a|}{2}+a<x$ $\implies$ $-\dfrac{3|a|}{2}=-\dfrac{|a|}{2}-|a|\le-\dfrac{|a|}{2} + a< x$
which implies that $\dfrac{1}{x}<-\dfrac{2}{3|a|}<\dfrac{2}{3|a|}<\dfrac{2}{|a|}.$Thus verifying that $\dfrac{1}{|x|}<\dfrac{2}{|a|}$.
Now using this information, I am to show that in general, $f(x)=1/x$ approaches $1/a$ for $x$ near $a$. My approach so far: $\biggl|\dfrac{1}{x} - \dfrac{1}{a}\biggr|=\biggl|\dfrac{a-x}{ax}\biggr|=\dfrac{1}{a}\cdot\dfrac{1}{|x|}\cdot |x-a|<\dfrac{1}{a}\cdot \dfrac{2}{|a|}\cdot |x-a|$ and so requiring that $|x-a|<\dfrac{|a|\cdot a}{2}\epsilon$ gives us our desired result.
What is the problem with allowing negative values of $x$?
If $a > 0$ but an interval $( a+\delta, a-\delta)$ contains some $x < 0$, then it also contains $0$, where $f$ isn't defined.
Also, we can do it without looking at multiple cases for $a$. Assume that $|x-a| \le \frac{|a|}2$. Then $$|a|-|x| \le |x-a| \le \frac{|a|}2 \implies |x| \ge \frac{|a|}2$$ In particular $x \ne 0$. We have
$$\left|\frac1x-\frac1a\right| = \left|\frac{a-x}{xa}\right| = \frac{1}{|x|}\cdot \frac1{|a|}\cdot |x-a| \le \frac{2}{|a|}\cdot \frac1{|a|}\cdot|x-a| \le \frac{2|x-a|}{|a|^2}$$ Hence for arbitrary $\varepsilon > 0$ if $|x-a| < \min\left\{\frac{|a|}2, \frac{\varepsilon |a|^2}{2}\right\}$ we get $\left|\frac1x-\frac1a\right| < \varepsilon$.