Find the limit of function using an epilson delta proof Find $\lim\limits_{x \to 4} \frac{\sqrt{x}-2}{x-4}$ using an epilson delta proof.
I multiplied and divided by the rational conjugate to get
$\frac{\sqrt{x}+2}{x-4}=\frac{1}{\sqrt{x}+2}$  So the limit is probably $\frac{1}{4}$. 
I can't seem to figure out how to turn $|\frac{1}{\sqrt{x}+2}-\frac{1}{4}|$ into something of the form $|x-4||f(x)|$ where f(x) is some function that I can then find the maximum value over [3,5] to find the value of delta. 
 A: 
(answer to the edited question)

We claim that $$\lim_{x\to 4}\frac{1}{\sqrt{x}+2}=\frac{1}{4}.$$
Note that $\sqrt{x}+2>1$ and so
$$0<\frac{1}{\sqrt{x}+2}<1. \tag 1$$ 
Also, $$x-4=(\sqrt{x}+2)(\sqrt{x}-2).\tag 2$$
Now,
$$\begin{align}
\bigg|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\bigg|&=\bigg|\frac{2-\sqrt{x}}{4(\sqrt{x}+2)}\bigg|\\
&=\frac{|2-\sqrt{x}|}{|4(\sqrt{x}+2)|}\\
&=\frac{|\sqrt{x}-2|}{4(\sqrt{x}+2)}\\
&=\frac{1}{4}\cdot\frac{1}{\sqrt{x}+2}\cdot |\sqrt{x}-2|\quad\text{then use $(1)$ to get}\\
&<\frac{|\sqrt{x}-2|}{4}.
\end{align}$$ 
But using $(2)$ and $(1)$, we get
$$|\sqrt{x}-2|=\bigg|\frac{x-4}{\sqrt{x}+2}\bigg|<|x-4|.$$ Hence,
$$\bigg|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\bigg|<\frac{|x-4|}{4}.$$ 
Finally, we pick $\epsilon>0$ and take $\delta=4\epsilon$ (you can also take $\delta$ such that $0<\delta<4\epsilon$). If $0<|x-4|<\delta$ then $$\bigg|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\bigg|<\frac{|x-4|}{4}<\frac{\delta}{4}=\epsilon.$$ This proves that 
$$\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}=\lim_{x\to 4}\frac{1}{\sqrt{x}+2}=\frac{1}{4}.$$
A: Hint: (For $x>4$)
$\forall N>0$, $\exists\delta>0$ such that if $|x-4|<\delta$ then $$\Big|\frac{\sqrt{x}+2}{x-4}\Big|=\frac{\sqrt{x}+2}{\big|x-4\big|}>\dfrac{2}{\delta}=N$$
