Limit of $\frac{x+3}{1+\sqrt x}$. 
Prove that $\lim\limits_{x\to1}\frac{x+3}{1+\sqrt x}=2$.

This problem has a solution on this site, but I can't seem to get the same answer is them. Their solution is to let $\delta=\min\{1,\epsilon\}$, but my method suggested I should use $\delta=\min\{1,\frac{\epsilon}{\sqrt 2-1}\}$.
Here is my question: does it matter? Clearly you can always round down your delta value, so is it just that I went through the trouble to find the largest possible $\delta$ when it suffices to use $\delta=\min\{1,\epsilon\}$?

Here is the outline of my proof
begin with the fact that we need $$\left|\frac{x+3}{1+\sqrt x}-2\right|<\epsilon$$
this (check the link) leads to $$\left|\frac{(x-1)(\sqrt x-1)}{(1+\sqrt x)^2}\right|<\epsilon$$
or $$|x-1||\sqrt x-1|\left|\frac{1}{(1+\sqrt x)^2}\right|<\epsilon$$
Ok so here we will pick $\delta\le1$, and I differ from the answer provided.
$|x-1|<1\implies-1<x-1<1\implies 0<x<2\implies \sqrt x<\sqrt 2$
Then we have both $\sqrt x-1<\sqrt 2-1$ the fraction $\frac{1}{(1+\sqrt x)^2}$ is always less than 1.
Now we want $|x-1|(\sqrt 2-1)$ to be less than $\epsilon$, so we pick $|x-1|<\frac{\epsilon}{\sqrt2-1}$.

As I mentioned, it seems to me that if my answer, $\delta=\min\{1, \frac{\epsilon}{\sqrt 2-1}\}$, works, then $\delta=\min\{1,\epsilon\}$ would also work.
In general should I find the simplest $\delta$?

You are correct that $|x-1|<1\implies\sqrt{x}-1<\sqrt{2}-1$, but
you can't conclude from this that $\lvert\sqrt{x}-1\rvert<\sqrt{2}-1$.  
For example, $x=\frac{1}{4}$ gives $\lvert\sqrt{x}-1\rvert=\frac{1}{2}>\sqrt{2}-1$
