<s>[INST] 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*}
 [/INST] Note that\begin{align}\frac{x^2-9}{x^3-3x^2-2x+6}-\frac67&=\frac{-6 x^2+7 x+33}{7 \left(x^2-2\right)}\\&=\frac{-6x-11}{7(x^2-2)}(x-3).\end{align}So, take $\delta_1>0$ such that $|x-3|<\delta_1\implies|x-3|<1\iff2<x<4$. Then$$7(x^2-2)>7(2^2-2)=14\quad\text{and}\quad|-6x-11|=|6x+11|<35.$$ Therefore$$|x-3|<\delta_1\implies\left|\frac{x^2-9}{x^3-3x^2-2x+6}-\frac67\right|<\frac{35}{14}|x-3|.$$Now, take $\delta>0$ such that $\delta\leqslant\delta_1$ and also that $\delta<\frac{14}{35}\varepsilon$. Then$$|x-3|<\delta\implies\left|\frac{x^2-9}{x^3-3x^2-2x+6}-\frac67\right|<\frac{35}{14}\times\frac{14}{35}\varepsilon=\varepsilon.$$
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