$\lim_{x \rightarrow 4} \frac{\sqrt{x}-2}{x-4}$. Did I choose my $\delta$ Properly? Preliminaries:
We know that, the limit as $ x \rightarrow a $ of a function  $ f(x) $ is $ L$ if for every $ \epsilon > 0 $ there exist a  $ \delta > 0 $ such that $ 0 <|x-a|<\delta \implies |f(x) - L|<\epsilon$. Our function is $ f(x) = \frac{\sqrt{x} - 2}{x - 4}$. 
Note that 
$$
 \lim_{x \rightarrow 4}  \frac{\sqrt{x} - 2}{x - 4} 
$$ 
cannot be solved as it is, for replacing $ x $ for $ 4 $ will render an undetermined answer. In order to fix this, we can rationalize the denominator by performing: 
 $$
  \frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{(x-4)}{(x-4)(\sqrt{x}+2)} = \frac{1}{\sqrt{x}+2}
$$ 
Let the rationalized expression be $ g(x) $. as $ g(x) = f(x) $   we can compute the limit: 
$$  
\lim_{x \rightarrow 4}\frac{1}{\sqrt{x}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{2+2} = \frac{1}{4}
$$
The Process for Choosing the Delta: 
We shall say, that the limit as $ x \rightarrow 4 $ of $ \frac{1}{\sqrt{x}+2}$ is $\frac{1}{4}$ if for every $ \epsilon > 0 $ there exist a $ \delta > 0$ such that $ 0 < | x - 4| < \delta \implies \left|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\right|<\epsilon$ 
Thanks to the property of triangle inequality, we have that: 
$$
\left|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\right| \leq \left| \frac{1}{\sqrt{x}+2}\right| - \left|\frac{1}{4}\right|
$$
For instance, it is sufficient for 
$$
\left| \frac{1}{\sqrt{x}+2}\right| - \frac{1}{4} < \epsilon
$$
which is equivalent to
$$
\left|\frac{1}{\sqrt{x}+2}\right| < \frac{4\epsilon+1}{4}
$$
Now, as per the definition of absolute value inequality, we have that: 
$$
-\frac{4\epsilon + 1}{4} < \frac{1}{\sqrt{x}+2} < \frac{4\epsilon + 1}{4} 
$$
By raising the expression to the power of minus one on each "side of the inequality" we can argue that,
$$
-\frac{4}{4\epsilon +1} < \frac{\sqrt{x}+2}{1} < \frac{4}{4\epsilon+1}
$$
which is equivalent to 
$$
\left|\sqrt{x}+2\right| < \frac{4}{4\epsilon+1} 
$$
by using the triangle inequality once again, we can argue that  $\left|\sqrt{x}+2\right| \leq \left| \sqrt{x} \right| + |2|$
which means that it suffices for: 
$$
\left|\sqrt{x}\right| + 2 < \frac{4}{4\epsilon + 1}
$$
$$
\left| \sqrt{x} \right| < \frac{4}{4\epsilon+1} - 2
$$
$$
\left|\sqrt{x}\right| < \frac{4-2(4\epsilon +1)}{4\epsilon+1}
$$
If we square both sides of the equation, we shall have
$$
\left|\sqrt{x}\right|^2 = \left|\sqrt{x}\right| \cdot \left|\sqrt{x}\right| = \left| x \right| < \left(\frac{4-2(4\epsilon +1)}{4\epsilon+1}\right)^2
$$
Now, using once again the triangle inequality, we have that
$$
\left|x-4\right| \leq \left|x\right| + \left|- 4\right|
$$
for instance, if 
$$\left|x\right|  + \left|- 4\right| < \left(\frac{4-2(4\epsilon +1)}{4\epsilon+1}\right)^2 + \left|-4\right| = \left(\frac{4-2(4\epsilon +1)}{4\epsilon+1}\right)^2 + 4$$
we will have
$$
\left| x-4 \right| <  \left(\frac{4-2(4\epsilon +1)}{4\epsilon+1}\right)^2 + 4$$
For instance, if we choose
$$ \delta =\left(\frac{4-2(4\epsilon +1)}{4\epsilon+1}\right)^2 + 4$$ 
The Limit will exist.
question: Did I choose my Delta Correctly, and for instance is my proof correct?


Thanks to the property of triangle inequality, we have that: 
  $$
\left|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\right| \leq \left| \frac{1}{\sqrt{x}+2}\right| \color{red}{\mathbf{-}} \left|\frac{1}{4}\right|
$$

This is not correct.
Instead, you could rewrite:
$$\left|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\right|=\frac{1}{4}\left|\frac{2-\sqrt{x}}{2+\sqrt{x}}\right|=\frac{1}{4}\left|\frac{4-x}{\left(2+\sqrt{x}\right)^2}\right|=\frac{1}{4}\frac{\left|x-4\right|}{\left(2+\sqrt{x}\right)^2} \le \frac{\left|x-4\right|}{16}$$
And since $x \to 4$, you can make this arbitrarily small.
