Prove $f(x)=\sqrt{2x-6}$ is continuous at $x=4$ by using precise definition. Please help me get the answer to this question.
Prove $f(x)=\sqrt{2x-6}$ is continuous at $x=4$ by using precise definition. ($\epsilon-\delta$ definition of limits.)

There is no problem with "reo's" work, he/she gave you a good sketch of the background work. Now we will proceed in the forward argument which somehow where we "magically" know what $\delta$ has to be. 
Proof: Given $\epsilon >0$ choose $\delta $= min{$1,\frac{\epsilon}{\sqrt{2}}$}. Then, $|f(x)-f(4)| = \sqrt{2}|\frac{x-4}{\sqrt{x-3}+1}|$. 
Suppose $|x-4|<\delta \Rightarrow \sqrt{2}|\frac{x-4}{\sqrt{x-3}+1}|< \sqrt{2}\delta = \sqrt{2}(\frac{\epsilon}{\sqrt{2}})=\epsilon$.
$\textbf{Additional}$: Note: We can also have, $|x-4|< \delta \leq 2 \Rightarrow 2<x<6 \Rightarrow \frac{|x-4|}{|\sqrt{x-3}+1|}<\frac{\delta}{2}$. Thus, picking $\delta$=min{$2\epsilon,1$} works as well. We typically want to bound $x-a$ so that the denominator doesn't blow up.
