$\lim_{x \to 1} \frac{1}{x+2} = 1/3$ with $\varepsilon$-$\delta$ definition? I'm trying to use the $\varepsilon$-$\delta$ definition of a limit to prove that
$$
\lim_{x \to 1} \frac{1}{x+2} = 1/3.
$$
But I'm getting stuck on finding the correct $\delta$. Here is my try:
\begin{align*}
\lvert f(x) - L \rvert < \varepsilon \\
\lvert  \frac{1}{x+2} - \frac{1}{3} \rvert < \varepsilon \\
\lvert  \frac{x -1}{2 + x} \rvert < 3\varepsilon.
\end{align*}
And then I'm not really sure what to do. How do you proceed from here?

As the function is monotonic around $x=1$ (provided $\delta<3$), you can solve
$$\frac1{1+\delta+2}-\frac13=\pm\epsilon$$
which gives
$$\delta=\mp\frac{9\epsilon}{1\pm3\epsilon}.$$
Then take the smallest of the two absolute values,
$$\delta=\frac{9\epsilon}{1+3\epsilon},$$ which is the tightest possible.
Below, $\epsilon=\dfrac1{10}$ and $\delta=\dfrac9{13}$.

