Epsilon- Delta Proof $$
\lim_{x\to 1}  \frac{1}{x^2+2} = \frac{1}{3}
$$
I'm having a problem proving this using $\epsilon-\delta$ proofs. For some reason When I solve for $\epsilon$, I get a negative number. Since this value is supposed to equal $\delta$ and $\delta$ can't be negative I'm not sure how to move forward.
Thanks!
 A: Since$$\frac1{x^2+2}-\frac13=\frac{1-x^2}{3(x^2+2)}=-(x-1)\frac{x+1}{3(x^2+2)},$$the first thing to do is to choose $\delta'$ such that, when $|x-1|<\delta'$, $\left|-\frac{x+1}{3(x^2+1)}\right|$ is not large. Take $\delta'=\frac12$, for instance. Then $|x+1|<\frac52$ and $x^2+2>\frac94$. Therefore,$$\left|-\frac{x+1}{3(x^2+2)}\right|<\frac{10}{27}.$$So, take $\delta=\min\left\{\frac{27}{10}\varepsilon,\frac12\right\}$.
A: The problem is that “solving for $\epsilon$” is too restrictive in general.  Let's unpack @mfl's hint.
$$
    \left| \frac{1}{x^2+2} - \frac{1}{3} \right| 
    = \left|\frac{1-x^2}{x^2+2}\right|
    = \left|\frac{(x-1)(x+1)}{x^2+2}\right|
    = |x-1|\cdot |x+1| \cdot \frac{1}{x^2+2}
$$
We want to make this product smaller than $\epsilon$.  The only factor we can control directly is $|x-1|$; we can choose any $\delta$ we want to make that small.  The trick is to control the other factors based on the assumption that $|x-1|<\delta$.
One factor is easy: $x^2 \geq 0$ for all $x$, so $x^2 + 2 \geq 2$ for all $x$, so $\frac{1}{x^2+2} \leq \frac{1}{2}$ for all $x$.
To control $|x+1|$, notice that
\begin{gather*}
    |x-1| < \delta \implies -\delta < x-1 < \delta \\
\implies 1-\delta < x   < 1+\delta \\
\implies 2-\delta < x+1 < 2+\delta \\
\end{gather*}
If we know that $\delta \leq 1$, we can say $1 < x+1 < 3$, so $|x+1|<3$.
[We could use any positive number in place of $1$, but $1$ is the traditional choice.]
This would give us
$$
    \left| \frac{1}{x^2+2} - \frac{1}{3} \right|
    < \delta \cdot 3 \cdot \frac{1}{2} = \frac{3}{2}\delta
$$
If we know also that $\delta \leq \frac{2}{3}\epsilon$, we can conclude that the right-hand side above is $<\epsilon$, and that's exactly what we want.
We can choose $\delta$ as small as we need; how can we ensure that both $\delta \leq 1$ and $\delta \leq \frac{2}{3}\epsilon$?  By setting
$$
    \delta = \min\left\{1,\frac{2}{3}\epsilon\right\}
$$

This is not the end; you still need to write it up “forwards.”  But hopefully it does explain how to work around quotients and other factors.
A: For a given $\epsilon$, choose $\delta$ such that
$$\frac{\delta(\delta+2)}{2}<\epsilon.$$
Then, if 
$$|x-1|<\delta$$
then
$$|x|<\delta+1$$
and so
$$|x+1|\leq|x|+1<\delta+2.$$
So (from the hint you received in the comments)
$$\left|\frac{1}{x^2+2}-\frac{1}{3}\right|\leq\frac{|1-x^2|}{2}<\epsilon.$$
A: When $x, y\in [0, 2]$, one has
$$\left|\frac{1}{x^2+2} - \frac{1}{y^2+2}\right|
= \left|y-x\right| \frac{y+x}{(x^2+2)(y^2+2)}\le |y-x|\frac{4}{4} \le |y-x|
$$
Take $y = 1$ and $\delta = \epsilon$.
A: For every $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that
$|x-1| \lt \delta$ implies $|\dfrac {1}{x^2+2} -1/3| \lt \epsilon$.
$F(x):=$
$|\dfrac {1}{x^2+2} -1/3| = |\dfrac {1-x^2}{x^2+2}| =$
$|\dfrac {(1-x)(1+x)}{x^2+1}|.$
Let $0\lt x \lt 2$, then $1 \lt |1+x| \lt 3,$
and $|x-1| \lt 1.$
Choose $\delta = \min ( (2/3) \epsilon, 1)$. 
$|x-1| \lt \delta$  implies
$|F(x)| = |\dfrac{(1-x)(1+x)}{x^2+2}| \lt$
$|\dfrac{(1-x)(1+x)}{2}|\lt$
$(1/2)3\delta =\epsilon$.
Note: Quite straight forward, only the term $|1+x|$ needs extra attention.
