Having trouble proving epsilon deltas of limits with roots I am having a big issue proving this equation... Specifically at the part that involves the roots. I can't seem to get $\sqrt{2x-1}/\sqrt{x-3} - \sqrt{7}$ to look like $x-4$ in terms of epsilon. In other words I don't know how I can algebraically manipulate the root part of this proof to make it look like the $x-4$ part of it.

 A: For the calculation, rewrite
$$\frac{\sqrt{2x-1}}{\sqrt{x-3}}-\sqrt{7}$$
as 
$$\frac{\sqrt{2x-1}-\sqrt{7}\sqrt{x-3}} {\sqrt{x-3}}.$$
Multiply top and bottom by
$$\sqrt{2x-1}+\sqrt{7}\sqrt{x-3}.$$
On top you end up with $-5(x-4)$. Nice!
Now you need to ensure the new bottom is not too small. So first specify that $\delta\lt 3/4$. That ensures that $\sqrt{x-3}\gt \frac{1}{2}$. 
A: Given any $\epsilon>0$, let $\delta = \min\{\epsilon/A, B\}$, where $A,B$ are some positive constants that we're going to figure out later. Then if $0<|x-4|<\delta$, observe that:
\begin{align*}
\left| \frac{\sqrt{2x-1}}{\sqrt{x-3}}-\sqrt{7} \right|
&= \left| \frac{\sqrt{2x-1}-\sqrt{7}\sqrt{x-3}} {\sqrt{x-3}} \right| \\
&= \left| \frac{\sqrt{2x-1}-\sqrt{7}\sqrt{x-3}} {\sqrt{x-3}} \cdot \frac{\sqrt{2x-1}+\sqrt{7}\sqrt{x-3}}{\sqrt{2x-1}+\sqrt{7}\sqrt{x-3}} \right| \\
&= \left| \frac{(2x-1)-7(x-3)} {\sqrt{x-3}(\sqrt{2x-1}+\sqrt{7}\sqrt{x-3})} \right| \\
&= \left| \frac{-5(x-4)} {\sqrt{x-3}(\sqrt{2x-1}+\sqrt{7}\sqrt{x-3})} \right| \\
&\leq \left| \frac{-5(x-4)} {\sqrt{x-3}(0+\sqrt{7}\sqrt{x-3})} \right| \qquad\text{since }\sqrt{2x-1}\geq0\\
&= \left| \frac{-5(x-4)} {\sqrt{7}(x-3)} \right|\\
&= \left| \frac{-5}{\sqrt 7}  \right| \cdot \frac{1}{|x-3|} \cdot |x-4|\\
&=  \frac{5}{\sqrt 7} \cdot \frac{1}{|x-3|} \cdot |x-4|\\
&< \frac{5}{\sqrt 7} \cdot \frac{1}{C} \cdot \frac \epsilon A \qquad \text{since } |x-4|<\delta \leq \epsilon/A\\
&= \epsilon \\
\end{align*}
The hard part is figuring out $A,B,C$. Somehow, we need to choose $B$ such that $|x-4|<\delta \leq B \implies |x-3|>C$ for some $C>0$ so that $\dfrac{1}{|x-3|} < \dfrac1C$ (this justifies the second to last step). Once we've done that, it suffices to let $A = \dfrac{5}{C\sqrt 7}$ (this justifies the last step).
With this in mind, suppose we choose $B=1/2$. Then we have:
\begin{align*}
|x-4| &< 1/2 \\
-1/2 < x-4 &< 1/2 \\
1/2 < x-3 &< 3/2 \\
\end{align*}
Hence, we have $|x-3| > 1/2$, so we obtain $C=1/2$. Thus, we may let $A = \dfrac{10}{\sqrt 7}$.
