Work So Far: I know the general definition of $\varepsilon-\delta$ continuity at a point $x_0$: $\forall \varepsilon > 0$ $\exists \delta > 0$ such that
$$|x-x_0| < \delta \implies |f(x)-f(x_0)| < \varepsilon$$
Of course, as usual, it is best to begin by setting out to find the necessary $\delta$. The starting point for my problem is thus
$$|x-1| < \delta \implies \left| \frac{1}{x+2} - \frac 1 3 \right| < \varepsilon$$
Through successive manipulations, we see that
\begin{align*} \left| \frac{1}{x+2} - \frac 1 3 \right| &= \left| \frac{3}{3(x+2)} - \frac{x+2}{3(x+2)} \right| \\ &= \left| \frac{3-(x+2)}{3(x+2)} \right| \\ &= \left| \frac{-x+1}{3(x+2)} \right| \\ &= \frac{|x-1|}{3|x+2|} \\ &< \varepsilon \end{align*}
Clearly the assumption $|x-1| < \delta$ is meant to be used here; however, $\delta$ shouldn't be a function of $x$, so I need to do something with the $|x+2|$ in the denominator...
The Question: How do I proceed from here? (Preferably using different functions as examples because this is a homework problem and I want to ultimately do it myself.)
From what I understand, I imagine this is a case where $\delta$ should be chosen to be the minimum of a pair of values, one (usually) a constant, and the other to be a multiple of $\varepsilon$.
However when I've seen such problems in the past - e.g. as examples in texts, or as MSE posts when searching for an answer to this - what to use for $\delta$ has often felt like it's just been "pulled out of thin air," so to speak. A thorough explanation as to how one derives those values (again, in particular alongside examples) would be greatly appreciated, because this is something that's been bugging me for a long time and I've never had a proper answer.
And thanks for the insights you can offer!
Since we're interested in $x\to 1$, then we can find a way to replace $|x+2|$ by a convenient constant. If $|x-1| < 1$, for example, then $|x+2| > 2$.
Using this fact, $$\frac{|x-1|}{3|x+2|} < \frac{|x-1|}6.$$
Now, if $|x-1| < \delta$, then we would have $$\frac{|x-1|}{3|x+2|} < \frac\delta6.$$
Wouldn't it be convenient if $\frac\delta6$ were equal to $\varepsilon$? In other words, if $\delta=6\varepsilon$?
So this is where you get to rewrite your proof, pulling $\delta = 6\varepsilon$ seemingly out of thin air...