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!

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.
