Prove that $\lim _{x\rightarrow 4}\sqrt {x}=2$ using the limit definition. My proof. Let $\varepsilon >0$. Pick $\delta=5\varepsilon$.
If $\left| x-4\right| < \delta$ for $x\in\mathbb{R} ^{\geq 0}$ then $\left| \sqrt {x}-2\right| =\dfrac {\left| x-2\right| } {\left| \sqrt {x}+2\right| }=\dfrac {\left| x-2\right| } {\left| \sqrt {x}\right| +2}=\delta/5\leq \dfrac {5\varepsilon } {\varepsilon }=\varepsilon$.
Note that for $0<x<4$, $\left| \sqrt {x}+2\right| \leq \left| \sqrt {x}\right| +2\leq 5$.
Can you check my proof?
 A: Another very similar way that may add some clarity:
Let $\epsilon>0$ be given. We need to find $\delta>0$ such that if $|x-4|<\delta$, then $|\sqrt{x}-2|<\epsilon$. Note that
$$
|\sqrt{x}-2|=\left|\sqrt{x}-2\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}\right|=\left|\frac{x-4}{\sqrt{x}+2}\right|=\frac{|x-4|}{\sqrt{x}+2}.
$$
To make the above chain of reasoning useful, we may experiment with values of $\delta$ (often starting with $\delta=1$ although sometimes this may prove to be problematic...occasionally, one needs to choose smaller values of delta to begin with, but $\delta=1$ will do just fine here). We let $\delta=1$ and note that
$$
|x-4|<1\Leftrightarrow -1<x-4<1\Leftrightarrow 3<x<5\Leftrightarrow \sqrt{3}+2<\sqrt{x}+2<\sqrt{5}+2.
$$
This suggest we should choose $\delta=\min\{1,\epsilon(\sqrt{3}+2)\}$. If we do this, then we note that
$$
|\sqrt{x}-2|=\frac{|x-4|}{\sqrt{x}+2}<\frac{|x-4|}{\sqrt{3}+2}<\frac{\epsilon(\sqrt{3}+2)}{\sqrt{3}+2}=\epsilon,
$$
as desired.
A: For the epsilon that you have, first make delta to be small, and smaller than $1$. So $|x-4| < 1 \implies 4-|x|=|4| - |x| < |x-4| <1\implies |x| > 3\implies \sqrt{x}+2 > \sqrt{3}+2\implies |\sqrt{x}-2| < \dfrac{|x-4|}{\sqrt{3}+2}< \dfrac{\delta}{\sqrt{3}+2}< \epsilon$, if $\delta < \epsilon(\sqrt{3}+2)$. Thus you can figure out how to take the delta. Simply make $0 < \delta < \text{min}(1,\epsilon(\sqrt{3}+2))$.
