Prove by limit definition $\lim _{x\to 3}\left(\frac{x^2-9}{x^3-3x^2-2x+6}\right)=\frac{6}{7}$. Prove, using the limit definition, that
$\lim _{x\to 3}\left(\frac{x^2-9}{x^3-3x^2-2x+6}\right)=\frac{6}{7}$.
I tried to do it this way, but I can’t move forward. Can anyone help me how to continue?
By definition we have
$\forall \epsilon >0,\exists \delta >0\::\:0<\left|x-a\right|<\delta \:\Rightarrow \left|f\left(x\right)-L\right|<\epsilon \Leftrightarrow a-\delta<x<a+\delta \Rightarrow L-\epsilon<f(x)<L+\epsilon.$
Proof:
\begin{eqnarray*}
 3-\delta <x<3+\delta \:\Rightarrow \:\frac{6}{7}-\epsilon <\:f\left(x\right)<\frac{6}{7}+\epsilon &\:\Leftrightarrow& \:\frac{6}{7}-\epsilon <\frac{\left(x-3\right)\left(x+3\right)}{x^2\left(x-3\right)+2\left(x-3\right)}<\frac{6}{7}+\epsilon\\ &\:\Leftrightarrow& \:\frac{6}{7}-\epsilon <\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x^2-2\right)}<\frac{6}{7}+\epsilon\\ &\:\Leftrightarrow& \:\frac{6}{7}-\epsilon <\frac{\left(x+3\right)}{\left(x^2-2\right)}<\frac{6}{7}+\epsilon \\
 &\:\Leftrightarrow& \:\:?
\end{eqnarray*}

If you want to avoid complications, just replace $x\to 3$ by $x=u+3$ with $u\to 0$.
Then $\left|f(x)-\dfrac 67\right|=\underbrace{\left|\dfrac{6u+29}{7(u^2+6u+7)}\right|}_\text{bounded}\times\underbrace{|u|}_{<\varepsilon}\to 0$
It is bounded because $|u|<1\implies \begin{cases}|6u+29|<35\\|u^2+6u+7|\ge |6u+7|\ge\big|7-6|u|\big|>1\end{cases}$
