How to prove this limit via definition?

Question

I would like to know how to prove this limit via the epsilon delta definition:

$$\lim\limits_{x\to3}\frac{x+1}{(x-3)^2}=\infty$$

Answer 1

$\forall M >0, \exists \delta > 0 : |x-3|<\delta \implies |\frac {x+1}{(x-3)^2}|> M$

Lets force $\delta < 1$ which will force $x+1>2$

$|\frac {x+1}{(x-3)^2}| > \frac {2}{\delta^2} > M$

When $\delta< \min (\sqrt {\frac {2}{M}}, 1),\frac {x+1}{(x-3)^2}> M$