Use of definition of limit to prove $\lim_{x→2} \dfrac{x}{x^2-2}=1$ I know that by definition I have to prove that $$\lim_{x→2}\dfrac{x}{x^2−2}=1⟺∀ϵ>0,∃δ>0,\, 0<|x−2|<δ⟹\left|\frac{x}{x^2−2}−1\right|<ϵ.$$
I have:
$\left|\dfrac{-x^2+x+2}{x^2-2}\right|= \left|\dfrac{-(x^2-x-2)}{x^2-2}\right|= \left|\dfrac{-(x^2-x+1-3)}{x^2-2}\right|=$
$\left|\dfrac{-(x-1)^2+3}{x^2-2}\right|$.
So, I don't know how to continue.

Just use the definition: you need to verify that
$$\Bigg|\frac{x}{x^2-2}-1\Bigg| < \epsilon$$
gives you a neighbourhood of $2$.
This means to solve
$$\begin{cases}
\dfrac{x}{x^2-2}-1 < \epsilon \\\\
\dfrac{x}{x^2-2}-1 > -\epsilon
\end{cases}
$$
Solve this and you will get
$$\sqrt{\dfrac{2\epsilon-4}{\epsilon-1}} < x < \sqrt{\dfrac{-2\epsilon-4}{-\epsilon-1}}$$
Which, being $\epsilon$ small ad libitum, gives you $2 < x < 2$.
