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.
On
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 $|d| < \delta = \min(1, \dfrac{1}{4} \varepsilon )$ $$\left| d \right| \left| \dfrac{d+3}{d^2+d+2} \right| < \dfrac{1}{4} \varepsilon \cdot \dfrac{4}{1} = \varepsilon.$$
On
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 $2 < x < 2$.
Note that $-(x^2-x-2)=-(x-2)(x+1)$ and that therefore$$\frac x{x^2-2}-1=-(x-2)\frac{x+1}{x^2-2}.$$Besides, if $|x-2|<\frac12$, then $\frac32<x<\frac52$ and therefore $\frac14<x^2-2<\frac{17}4$; in particular, $|x^2-2|>\frac14$. On the other hand, $\frac52<x+1<\frac72$; in particular, $|x+1|<\frac72$. So,\begin{align}\left|-(x-2)\frac{x+1}{x^2-2}\right|&=|x-2|\frac{|x+1|}{|x^2-2|}\\&<|x-2|\frac{7/2}{1/4}\\&=14|x-2|.\end{align}Therefore, given $\varepsilon>0$, take $\delta=\min\left\{\frac\varepsilon{14},\frac12\right\}$, and then$$|x-2|<\delta\implies\left|\frac x{x^2-2}-1\right|<\varepsilon.$$