Trying to pick up some math courses these days and gotten to epsilon-delta proofs in the calculus class. Working on some text-book problems now and since the text-book do these proofs slightly different from what I'm used to from the time I had this many, many years ago I'm not sure if my reasoning is correct.
As title says, problem at hand is $\lim_{x\to4}\sqrt{x}=2$. The approach used in the text book is to substitute $h=x-a$ e.g. $0<|x-a|<\delta=0<|h|<\delta$, and in the calculation of $|f(x)-b|<\epsilon$ use $x=h+a$.
I've tried the following:
$|f(x)-2|=|\sqrt{x}-2|=\left|\frac{x-4}{\sqrt{x}+2}\right|=\left|\frac{h}{\sqrt{h+4}+2}\right|=|h|\left|\frac{1}{\sqrt{h+4}+2}\right|$.
Now, my reasoning is that for $h>0$, $\sqrt{h+4}+2>\sqrt{4}+2=4$. For $h=0$ we obviously get the desired result, hence we can select $\delta=\min\left(1,4\epsilon\right)$, getting $|f(x)-2|=|h|\left|\frac{1}{\sqrt{h+4}+2}\right|<4\epsilon\cdot1/4=\epsilon$
I did some searching online, and I saw that someone has asked about the same problem at epsilon-delta proof of $\lim_{x \to 4} \sqrt{x} = 2$, but here it's done in a different manner and with a different result. In one of the comments in that question it is correctly stated that $\delta$ isn't unique; you only have to find a value that is good enough. Thus, my question basically is if my solution is good enough. If I get this right, I could also end up on the same result by saying $\sqrt{h+4}+2>\sqrt{4}+2>2>1$ and then select $\delta=\epsilon$ (or $2\epsilon$ if I quit at $\sqrt{4}+2>2$, given that the reasoning is correct at all).
let $\epsilon$ be a real$>0$.
for $x\in \mathbb R^+$
$$|\sqrt{x}-2|=\frac{|x-4|}{\sqrt{x}+2}$$
$$\leq \frac{|x-4|}{2}$$
cause $\sqrt{x}+2\geq2$.
so, if $|x-4|<2\epsilon$, then we will be
sure that $|\sqrt{x}-2|<\epsilon$.
thus we will take $\delta=2\epsilon$.