Prove $\lim_{x \to 2} \frac{ x(x^2-1) }{x+3}=6/5$. Prove by $\epsilon$-$\delta$ definition that $\displaystyle\lim_{x \to 2}  \frac{ x(x^2-1) }{x+3}=\dfrac{6}{5}$.
I know this is simple but I am stuck, If we assume $\delta <1$ and $|x-2|< \delta$  we can get rid of the factor $x-2$ in the following expression $$\dfrac{ x(x^2-1) }{x+3}-\dfrac{6}5=\dfrac{(x-2)(5x^2+10x+9)}{5x+15}$$
but I don't know how to proceed.

From here
$$\left|\frac{ x(x^2-1) }{x+3}-6/5\right|=\left|\frac{(x-2)(5x^2+10x+9)}{5x+15}\right|$$
the trick is consider wlog $|x-2|<1 \iff 1<x<3$ and therefore
$$\left|\frac{(x-2)(5x^2+10x+9)}{5x+15}\right|=|x-2|\left|\frac{5x^2+10x+9}{5x+15}\right|\le \frac{84}{20}|x-2|\le5|x-2|$$
therefore it suffices to take $\delta =\frac{\epsilon}5$ to fulfill the definition, indeed in this case for any $x$ such that $0<|x-2|< \delta$
$$\left|\frac{ x(x^2-1) }{x+3}-6/5\right|\le 5|x-2|< 5 \cdot\frac \epsilon 5=\epsilon$$
