Real Analysis limits of functions delta epsilon proof Prove from first principles that
$f(x) = \displaystyle\frac{x^2-4}{x-4}$ approaches $-5$ as $x$ approaches $3$.
I am terrible at these proofs. I know we start like this
Fix $\epsilon > 0$. We need to find $\delta$ such that
$0 < |x-3| < \delta \implies |f(x)-(-5)| < \epsilon$
I simplified the abs so that i have 
$ \left|  \displaystyle\frac{(x+8)(x-3)}{x-4} \right|< \epsilon$ but I dont know where to go from here.
Cheers!

We start from the inequality you arrived at. So we want to show that by making $x$ close enough to $3$, we can make $ \left|  \displaystyle\frac{(x+8)(x-3)}{x-4} \right|< \epsilon$.
We can exercise control over $|x-3|$. But there is a potential problem with $ \left|  \displaystyle\frac{x+8}{x-4} \right|$. We must make sure this does not get too big,
First of all, we will insist that $\delta\lt \frac{1}{2}$. Why? That will confine $x+8$ to the interval $(10.5,11.5)$. So in particular we will have $|x+8|\lt 11.5$.
Also, $x-4$ will be confined to the interval $(-1.5,0.5)$. So $|x-4|$ will be greater than $0.5$.
Thus, if $\delta$ is chosen as above, we will have $ \left|  \displaystyle\frac{x+8}{x-4} \right|\lt 23$, and therefore we will have
$$\left|\frac{(x+8)(x-3)}{x-4} \right|< 23\delta.$$
Thus we can force $ \left|  \displaystyle\frac{(x+8)(x-3)}{x-4} \right|$ to have absolute value $\lt \epsilon$ by picking $\delta\lt \frac{1}{2}$ and $\delta\lt \frac{\epsilon}{23}$.
So let $\delta=\min\left(\frac{1}{2},\frac{\epsilon}{23}\right)$. If $|x-3|\lt \delta$, the desired inequality holds. 
