<s>[INST] 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!
 [/INST] 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.
</s>