<s>[INST] 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.
 [/INST] 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$$|x-2|<\delta\implies\left|\frac x{x^2-2}-1\right|<\varepsilon.$$
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