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.

Two points: 1) It's often much easier to deal with $x\to 0$ than $x\to c\neq 0$; and  2) you may want to avoid fractions whenever you can.
So let's denote $d=x-2$ and prove $\lim_{d\to 0} \dfrac{d+2}{(d+2)^2-2} = 1$.
$$
\left| \dfrac{d+2}{(d+2)^2-2} - 1 \right|=\left| \dfrac{d+2-d^2-4d-2}{d^2+4d+2} \right| = \left| d \right| \left| \dfrac{d+3}{d^2+d+2} \right|
$$
If $|d|<1, |d+3| \le |d| + 3<4$. Also note $|d^2+d+2|=\left(d+\dfrac{1}{2}\right)^2+\dfrac{7}{4} > 1$.
Therefore, if $|d| < \delta = \min(1, \dfrac{1}{4} \varepsilon )$
$$\left| d \right| \left| \dfrac{d+3}{d^2+d+2} \right| < \dfrac{1}{4} \varepsilon \cdot \dfrac{4}{1} = \varepsilon.$$
