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.

If you found the algebra a bit hard to decipher, maybe this will help.
Assume that $|x-13|<1$ just means that we assume that the distance between $x$ and 13 is less than one. $x$ is the input to your function, it's distance from the number 13 is less than 1. If you want to compute your limit, i.e.  what happens to $f(x)$ as $x\rightarrow13$, the distance between $x$ and 13 is going to be less than 1 eventually anyway, so this is not an unreasonable thing to do.
$|x-13|<1\Longleftrightarrow-1<x-13<1$ These two statements mean the same thing. It's easier to work with the equation to the right.
You treat $-1<x-13<1$ like an ordinary equation except instead of doing the same thing to both sides of the equality sign you do the same thing to the left, in the middle and to the right.

$\begin{array}{ccccc}
-1 & < & x-13 & < & 1\\
-1+13 & < & x-13+13 & < & 1+13\\
12 & < & x & < & 14\\
12-4 & < & x-4 & < & 14-4\\
8 & < & x-4 & < & 10\\
\sqrt{8} & < & \sqrt{x-4} & < & \sqrt{10}\\
\sqrt{8}+3 & < & \sqrt{x-4}+3 & < & \sqrt{10}+3\\
 & \text{flip} &  & \text{flip}\\
\frac{1}{\sqrt{8}+3} & > & \frac{1}{\sqrt{x-4}+3} & > & \frac{1}{\sqrt{10}+3}\\
\end{array}$

All these moves are legal. What we have so far is that if the length between your input $x$ and 13 is less than one, i.e $|x-13|<1$, then $\frac{1}{|\sqrt{x-4}+3|}  <  \frac{1}{\sqrt{8}+3}$.
Now comes the epsilon-delta part. Remember that the limit you want to find is basically the same as asking "what happens to $f(x)$ when my input $x$ gets really close to 13", so let's consider the case where $x$'s distance from 13 is less than 1 (this has to happen eventually). 

When $|x-13|<1$, we know from before that $|\sqrt{x-4}-3|=|x-13|\frac{1}{|\sqrt{x-4}+3|}$, so now we have $|\sqrt{x-4}-3|=\text{something less than 1}\cdot\text{something less than }\frac{1}{\sqrt{8}+3}<\frac{1}{\sqrt{8}+3}$

Let's consider when $|x-13|<(\sqrt{8}+3)\epsilon$:

We know from before that $|\sqrt{x-4}-3|=|x-13|\frac{1}{|\sqrt{x-4}+3|}$, so now we have $|\sqrt{x-4}-3|=\text{something less than }(\sqrt{8}+3)\epsilon\cdot\text{something less than }\frac{1}{\sqrt{8}+3}<(\sqrt{8}+3)\epsilon \frac{1}{\sqrt{8}+3}=\epsilon$

Now we choose $\delta=\min\{1,(\sqrt{8}+3)\epsilon\}$. Why? Because there is a flaw in the argument above! What if $\epsilon$ is greater than 1? Then we can't say that $\frac{1}{|\sqrt{x-4}+3|}  <  \frac{1}{\sqrt{8}+3}$ - but wait! if $\epsilon$ is greater than 1, we already have a $\delta$ that will satisfy our definition: 1.

Example:
If $\epsilon=2.2$, can you find a $\delta$ such that $|x-13|<\delta \Rightarrow |f(x)-3|<2.2?$ Yes, $|x-13|<1$, then $|\sqrt{x-4}-3|<\frac{1}{\sqrt{8}+3}\approx 0.17$

When $\epsilon$ is less than 1, then we know that $\frac{1}{\sqrt{x-4}+3}  <  \frac{1}{\sqrt{8}+3}$ and we use the $\delta$ we derived using the fact that $|x-13|<1$, namely $\delta=(\sqrt{8}+3)\epsilon$
Think about after $\epsilon=2.2$,  when someone tries to pass me an $\epsilon$ like 0.001 so that $(\sqrt{8}+3)\epsilon<1$, then having chosen $|x-13|<\delta=\min\{1,(\sqrt{8}+3)\epsilon\}$, I can repeat my whole argument about how when $|x-13|<1$, then blah blah blah. I have a winning strategy against any player who tries to break me by passing me tiny or large epsilons.
