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:
$\left|x-5\right|\lt 4(x+5)\delta$

You should get rid of $x+5$, as it is a function of $x$.
In order to do this, we fix an arbitrary  delta say 1 and take minimum from 1 and a new delta which will be obtained.
More precisley  we already know that $|x-5|<\delta$.
Now put $\delta=1$ and $9<x+5<11$ is obtained.
Let us go back to our limit.
One may see that $4(x+5)(x-5)<4(x-5)\times 11=44(x-5)<44\delta$ and so we have to have $\delta<\frac{1}{44}\epsilon$.
Hence we can conclude that $\delta\leq min\{1,\frac{1}{44}\epsilon\}$
