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.

Let $\epsilon>0$. We want to pick a $\delta$ such that $|x-(-1)|=|x+1|<\delta\implies|f(x)-f(-1)|=|f(x)|<\epsilon$. 
Note that 
$$
|f(x)|=\bigg|\frac{3x^{2}-2x-5}{2x+3}\bigg|=\bigg|\frac{(3x-5)(x+1)}{2x+3}\bigg|=\bigg|\frac{3x-5}{2x+3}\bigg||x+1|.
$$
We are free top choose $\delta$ as we wish, and, in doing so, we will have a handle on making the $|x+1|$ term above as small as we want. To help us minimize $\big|\frac{3x-5}{2x+3}\big|$, let's see what happens if we would ensure $\delta <\frac{1}{3}$. 
If $|x+1|<\frac{1}{3}$, then $-\frac{1}{3}<x+1<\frac{1}{3}$. This implies that
$$
-1<3x+3<1\implies-9<3x-5<-7<9\implies|3x-5|<9
$$ 
and
$$
-\frac{2}{3}<2x+2<\frac{2}{3}\implies\frac{1}{3}<2x+3<\frac{5}{3}\implies\frac{1}{3}<|2x+3|.
$$
Thus, 
$$
\bigg|\frac{3x-5}{2x+3}\bigg|<\frac{9}{1/3}=27.
$$
So, if $\delta<\frac{1}{3}$, it follows that $|f(x)|<27|x+1|$. So, if in addition to this, $\delta<\frac{\epsilon}{27}$, we get that $|f(x)|<27\cdot\frac{\epsilon}{27}=\epsilon$.
Hence, we could choose $\delta=\min\{\frac{1}{3},\frac{\epsilon}{27}\}$. Then, if $|x+1|<\delta$, it follows that $|f(x)|<\epsilon$, as desired.
