<s>[INST] $\frac{x^2 -4}{x-4} \rightarrow -5$ as $x \rightarrow 3$, proving the limit from first principles. ($\delta - \epsilon$). This is a follow up to a previous post. I am sorry of this is considered spamming but I have made significant progress of the question since then. Hopefully I'm on the right track, an tips or advice on my methods would be much appreciated!
I have been asked to prove the limit of $\frac{x^2 -4}{x-4} \rightarrow -5$ as $x \rightarrow 3$. First I fix $\epsilon >0$ and find $\delta>0$ such that.
$$0<|x-3| <\delta \implies \bigg|\frac{x^2 -4}{x-4} + 5\bigg| < \epsilon$$
$$\impliedby \bigg|\frac{x^2+5x -24}{x-4}\bigg| < \epsilon$$
$$\impliedby \bigg|\frac{(x+8)(x-3)}{x-4}\bigg| < \epsilon$$
$$\impliedby \bigg|(x-3)\frac{x+8}{x-4}\bigg| < \epsilon$$
If $|x-3| \leq \frac{1}{2}$, then $[2\frac{1}{2},3\frac{1}{2}]$. Then $\frac{x+8}{x+4} \geq \frac{1}{2}$. Therefore $\frac{|x-3|}{2} < \epsilon \iff |x-3|<2\epsilon $.
Thus any $\delta \leq$ min$(\frac{1}{2}, 2\epsilon)$ has the required property.
Thanks for your time!
 [/INST] Your choice for $\delta$ does not work.
Let $\varepsilon = \frac{1}{4}$.
Then your formula gives $\delta = \frac{1}{2}$.
If $x = 3.4$ then
$$\quad \bigg|(x-3) \; \frac{x+8}{x-4}\bigg| = 7.6$$

If $x \in [2\frac{1}{2},3\frac{1}{2}]$ then
$$\quad |x+8| \le 12 $$
If $x \in [2\frac{1}{2},3\frac{1}{2}]$ then
$$\quad |x-4| \ge \frac{1}{2} $$
It follows that for every $x \in [2\frac{1}{2},3\frac{1}{2}]$ that
$$\quad \bigg|\frac{x+8}{x-4}\bigg| = \frac{|x+8|}{|x-4|} \le 24$$
To solve the OP's problem, for any $\varepsilon \gt 0$ challenge let 
$$ \delta = \text{min}\bigg(\frac{1}{2}, \frac{\varepsilon}{24}\bigg)$$
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