Scratch work for delta-epsilon proof for $\lim_{x \to 13} \sqrt{x-4} = 3$ 
Prove $\lim_{x \to 13} \sqrt{x-4} = 3$.
We need to show for all $E> 0$ there exists $D > 0$ such that
  if $0 < |x - 13| < D$, then $|\sqrt{x-4} - 3| < E$. Let me write D for delta and E for epsilon please.
Scratch-work here: $\color{darkred}{\text{ I understand the formal proof. Ergo just asking about this. } }$
Note that $|\sqrt{x-4} - 3| = |\sqrt{x-4} - 3|\dfrac{|\sqrt{x-4} + 3|}{|\sqrt{x-4} + 3|} = \dfrac{|x - 13|}{|\sqrt{x-4} + 3|}$.
  We can bound $|x - 13|$ for any choice of D, but we need a certain D to also bound $|\sqrt{x-4} + 3|$.

1. Ishfaaq's answer says 'the denominator will definitely cause trouble.' How?
Why won't bounding $|x - a|, |x - 13|$ bound the whole statements?
Beneath pp 83 of Spivak claims this too. How does '$|x + a|$ cause trouble'? 

Assume that $D < 1.$ 

2. What sanctioned assuming $D < 1$? How to know if $1$ is too small, too big? $\\$ In pp 83 of Spivak beneath, $D$ looks randomly chosen?
4. Ishfaaq's last paragraph. $\color{darkred}{\left|{x^2 - a^2}\right| \lt \delta^2 + \left|{}2a\right|\delta. \text{ But we can hardly equate this to ϵ... }}$
Why not? $d^2 + \left|{}2a\right|d = e \iff d(d + |2a|) = e$? 
Ishfaaq says $a = 1/2, e = d/2$ is a counterexample. But then $d^2 + 2|a|d < e \iff d^2 + d < d/2 \iff d(d - \frac{1}{2}) < 0 \iff 0 < d < 1/2.$ What founders? 

If this works, then any smaller D will also work.
  Then $|x - 13| < 1 \iff -1 < x - 13 < 1 \iff 12 < x < 14 \iff 8 < x - 4 < 10$
  $\implies \sqrt(8) < \sqrt{x - 4} < \sqrt(10) \iff  \color{blue}{\sqrt(8) + 3 < \sqrt{x - 4} + 3 < \sqrt(10) + 3} \\ \iff  0 < \color{blue}{\dfrac{1}{\sqrt(10) + 3} < \dfrac{1}{\sqrt{x - 4} + 3} < \dfrac{1}{\sqrt(8) + 3}} $
  $\implies \color{green}{\frac{1}{|\sqrt{x - 4} + 3|} < \frac{1}{\sqrt(8) + 3}} $
So then $|x - 13|\color{green}{\dfrac{1}{|\sqrt{x - 4} + 3} < \dfrac{1}{\sqrt(8) + 3}}|x - 13|$ and need this $ < E.$
  Thus, we need $|x - 13| < E * \color{green}{(\sqrt{8} + 3)}$. So this is our choice of D. But note that this only works when $D < 1$. Thus, we can take care of both conditions by choosing $D = \min\{1, E(\sqrt{8}+3)\}.$ █

$\color{darkred}{ \text{ 3. All this algebra fazed me. How to graph this to view all this algebra and $D$? Thanks. } }$



Also tried http://www.ocf.berkeley.edu/~yosenl/math/epsilon-delta.pdf.

Note that $\sqrt{x-4}+3\geqslant3$ hence $|\sqrt{x-4}-3|\leqslant\frac13|x-13|$. Thus, if $|x-13|\leqslant3\varepsilon$ then $|\sqrt{x-4}-3|\leqslant\varepsilon$.
Note that the function $x\mapsto\sqrt{x-4}$ is only defined on $x\geqslant4$ hence the condition $|x-13|\leqslant3\varepsilon$ guarantees that $\sqrt{x-4}$ exists, provided $\varepsilon\leqslant3$ (which is more than enough to conclude). A formula valid for every $\varepsilon$ would be $\delta=\min\{9,3\varepsilon\}$.
