Prove that $\lim \limits_{x \to 5}\left(4x^2-7\right)=93$ So I first need to determine the limit and then prove it: 

$\lim \limits_{x \to 5}\left(4x^2-7\right)$

So $L=93$
And thus $\left|f(x)-L\right|=\epsilon$ and $\left|x-c\right|=\delta$
Plugging in the values...
$\left|\left(4x^2-7\right)-93\right| \lt \epsilon$, which when factored gives $4(x+5)\left|x-5\right|\lt \epsilon$
So does this mean:
$\left|x-5\right|\lt 4(x+5)\delta$

the next step.
Find $\delta$ such that when $|x-5|<\delta, |4(x + 5)(x-5)| < \epsilon$
let $\delta = \min (1, \frac {\epsilon}{44})$
Why 44? 
Suppose $x-5 = \delta, |4(x + 5)(x-5)| = \delta \le 1 \implies |4(\delta + 10) \delta| = 4 \delta^2 + 40 \delta$
If $\delta \le 1$ then  $|4(x + 5)(x-5)| < 44 \delta$
$\delta = \min (1, \frac {\epsilon}{44}) \implies |4(x + 5)(x-5)| < \epsilon.$
