Using $\epsilon-\delta$ proof to prove continuity 
Use an $\epsilon-\delta$ proof to show that
  $f : R \setminus \left \{ \frac{-3}{2} \right \} \rightarrow R$ ,  $$f(x) = \frac{3x^2-2x-5}{2x+3}$$
  is continuous at $x = -1$

Hello there. Can anyone here help me with this? I know i need to show that $|x-l|<\delta$ implies $|f(x) - f(l)| < \epsilon$ but I don't know how to do it for this specific example.

Fix $\varepsilon>0$ and let $x\in\mathbb{R}\setminus\{-\frac{3}{2}\}$ such that $|x-(-1)|=|x+1|<\delta$ for $\delta>0$ to be chosen later. As $f(-1)=0$, we have :
\begin{align}
\left|f(x)-f(-1)\right|&=\left|\frac{3x^2-2x-5}{2x+3}-0\right|\\
&=\left|\frac{3x^2-2x-5}{2x+3}\right|.
\end{align}
Now observe that
\begin{align}
3x^2-2x-5&=3(x+1)\left(x-\frac{5}{3}\right),\quad\quad\quad\quad(*)
\end{align}
\begin{align}
\left|x-\frac{5}{3}\right|&=\left|x+1-1-\frac{5}{3}\right|\\
&\leq\left|x+1\right|+\left|1+\frac{5}{3}\right|\\
&<\delta+\frac{8}{3},\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(**)
\end{align}
and that (by the inverse triangle inequality)
\begin{align}
|2x+3|&\geq|2|x|-3|\\
&\geq3-2|x|\\
&>3-2(1+\delta)\\
&=1-2\delta\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(***)
\end{align}
so that :
\begin{align}
\left|\frac{3x^2-2x-5}{2x+3}\right|&=\frac{3|x+1|\left|x-\frac{5}{3}\right|}{|2x+3|}\\
&<\frac{3\delta\left(\delta+\frac{8}{3}\right)}{|1-2\delta|}.
\end{align}
We now choose $\delta<1/4$ for $1-2\delta>1/2>0$, and then
\begin{align}
\left|f(x)-f(-1)\right|&<6\delta\left(\delta+\frac{8}{3}\right).
\end{align}
Finally, if we choose $\varepsilon=6\delta\left(\delta+\frac{8}{3}\right)$ for any $\delta<1/4$, then we get
$$|x-(-1)|<\delta\quad\Longrightarrow\quad|f(x)-f(-1)|<\varepsilon$$
whence $f$ is continuous at $x=-1$.
