$\lim_{x \to 1} \frac{1}{x+2} = 1/3$ 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?
 A: So far you have deduced a chain of equivalences, and I will add one more to the chain:
\begin{align*}
& \lvert f(x) - L \rvert < \varepsilon \\
\iff & \lvert  \frac{1}{x+2} - \frac{1}{3} \rvert < \varepsilon \\
\iff & \lvert  \frac{x -1}{2 + x} \rvert < 3\varepsilon \\
\iff & \lvert x-1 \rvert < 3 \, \epsilon \, \lvert 2+x \rvert
\end{align*}
Reading this chain of equivalences backwards, you have proved so far that
$$\lvert x-1 \rvert < 3 \, \epsilon \, \lvert 2+x \rvert \implies \lvert f(x) - L \rvert < \varepsilon
$$
Your goal now is to find a value of $\delta>0$ such that
$$ \lvert x-1 \rvert < \delta \implies  \lvert x-1 \rvert < 3 \, \epsilon \, \lvert 2+x \rvert
$$
for then you will be done.
The first step towards finding $\delta$ is to work with the factor $\lvert 2+x \rvert$ separately. Your goal in this step is to find any lower bound to that factor. Intuitively, if $x$ is close to $1$ then it must be far from $-2$. To make this intuition rigorous, note that
$$\lvert x-1 \rvert < 1 \implies -1 < x-1 < +1 \implies 2 < x+2 < 4 \implies 2 < \lvert x+2 \rvert 
$$
Summarizing, we have found $\delta_1 = 1$ for which the following implications hold:
$$\lvert x-1 \rvert < \delta_1 \implies \lvert x-1 \rvert < 1 \implies 2 < \lvert 2+x \rvert \implies 6 \epsilon < 3 \epsilon \lvert 2+x \rvert
$$
Now we set $\delta_2 = 6 \epsilon$.
Assuming already that $\lvert x-1 \rvert < \delta_1 = 1$, if in addition we assume that $\lvert x-1 \rvert < \delta_2 = 6 \epsilon$ then it follows that
$$\lvert x-1 \rvert < \delta_2 = 6 \epsilon < 3 \epsilon \lvert 2+x \rvert
$$
And now we have found $\delta$, namely:
$$\delta = \min\{\delta_1,\delta_2\}
$$
Putting this altogether, assuming that $\lvert x-1 \rvert < \delta$ it follows that $\lvert x-1 \rvert < \delta_1 = 1$ and $\lvert x-1 \rvert < \delta_2 = 6 \epsilon$ and therefore
$$\lvert x-1| < 3 \epsilon \lvert x+2 \rvert
$$
This, as we have already shown, implies that $\lvert f(x)-L \rvert < \epsilon$.
A: Let $x=1+\delta$. Then we have:
$$\lim_\limits{\delta\to0} \left|\frac{1}{3+\delta}-\frac{1}{3}\right|$$
$$=\lim_\limits{\delta\to0} \left|\frac{\delta}{3+\delta}\right|$$
Letting $\delta=\epsilon$ means that for any $0<\epsilon<1$, the expression in the limit is less than $\epsilon$ as required.
A: The last inequality should be $|\frac {x-1} {2+x}| <3\epsilon$. Use the fact that $|2+x|=|3+(x-1)| \geq 3-|x-1|$. What you now need is $\frac {|x-1|} {3-|x-1|} <3\epsilon$. Cross multiply and will get the inequality in the form $|x-1| <\delta$ where $\delta=\frac {9\epsilon} {1+3\epsilon}$.
A: Firstly there should be $\left|\frac{1}{x+2} - \frac{1}{3}\right| = \left|\frac{x -1}{2 + x}\right|\cdot \frac{1}{3}$
After this, one more possible way is choose such neighbourhood of $x=1$, in which $|2+x|$ can be estimated from below. For example, when $\delta \lt 1$, then $|2+x| = 2+x \gt 1$. Now we have $\left|\frac{x -1}{2 + x}\right| \lt  |x-1|\lt \delta $. Are you able to finish from here, taking in account factor $\frac{1}{3}$?
A: 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 $\delta=\dfrac9{13}$.

