By using $\delta-\epsilon$ definition, show that $\lim_{x \rightarrow2}{\frac{3x^2+2x}{x^2-2}}=8$ By using $\epsilon-\delta$ definition, show that $$\lim_{x \rightarrow2}{\frac{3x^2+2x}{x^2-2}}=8$$ I have no idea on how to find $\delta$ here as I don't have $|x-2|$. Can anyone guide me ?

Let $u(x)=(3x^2+2x)/(x^2-2)$, then 
$$
u(x)-8=-(x-2)\frac{8+5x}{x^2-2}.
$$
If $3/2\lt x\lt5/2$, then $|8+5x|\lt8+5(5/2)=41/2$ and $x^2-2\gt(3/2)^2-2=1/4$. Hence, for every $3/2\lt x\lt5/2$, $$|u(x)-8|\leqslant|x-2|\cdot(41/2)\cdot4=82\cdot|x-2|. 
$$
Finally, for every $\varepsilon\leqslant41$, every $\delta\leqslant\varepsilon/82$ is suitable, and, for every $\varepsilon\geqslant41$, every $\delta\leqslant1/2$ is suitable. "Suitable" here means that $\delta$ is such that:
$$
\forall x,\quad|x-2|\leqslant\delta\implies|u(x)-8|\leqslant\varepsilon.
$$
