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.)
 A: 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.
A: We have $|x-4|< \delta$.
$|f(x)-f(4)|=|\sqrt {2x-6}-\sqrt {2}|=|\sqrt{2}(\sqrt{x-3}-1)|$
Now, multiplying the numerator and denominator by $(\sqrt{x-3}+1)$,
$=\sqrt{2} |\frac{x-4}{\sqrt{x-3}+1}$|
Now, ${\sqrt{x-3}+1}$ is always greater than or equal to $1$. Thus, $\frac{x-4}{\sqrt{x-3}+1}\leqslant (x-4)$.
Therefore, $|f(x)-f(4)|\leqslant \sqrt{2}|(x-4)| < \sqrt{2} \delta=\epsilon$
Thus, $f(x)$ is continuous at $x=4$.
A: Let $f:[3,\infty)\to \mathbb{R}$ be defined by
$$f(x)=\sqrt{2x-6}$$
for every $x\in[3,\infty)$. (It is important to define the function explicitly; that is, the domain is relevant for this proof.)
$f$ is continuous at $4$.
Proof. To prove $f$ is continuous at $4$, we will show that for each $\epsilon>0$, there exists $\delta>0$ such that for each $x\in[3,\infty)$, if $|x-4|<\delta$, then $|f(x)-f(4)|<\epsilon$.
Let $\epsilon>0$ be arbitrary. Notice for each $x\in[3,\infty)$,
$$|f(x)-f(4)|=\sqrt{2}|\sqrt{x-3}-1|=\sqrt{2}\left|\frac{x-4}{\sqrt{x-3}+1}\right|<\sqrt{2}|x-4|\tag{1}$$
Define $\delta\overset{\text{def}}{=}\dfrac{\epsilon}{\sqrt{2}}$.
It is important to understand why we defined $\delta$ this way (in relation to $\epsilon$).
Anyway, we continue with the proof.
For each $x\in[3,\infty)$, if $|x-4|<\delta$, then
$$|f(x)-f(4)|\overset{(1)}{<}\sqrt{2}|x-4|<\sqrt{2}\cdot\delta=\epsilon$$
Recall $\epsilon>0$ was arbitrary. Therefore for each $\epsilon>0$, there exists $\delta>0$ (namely $\epsilon/\sqrt{2}$), such that for each $x\in[3,\infty)$, if $|x-4|<\delta$, then $|f(x)-f(4)|<\epsilon$. Therefore $f$ is continuous at $4$.$\square$
