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!

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 $\delta = \epsilon$.
