# Use the definition of a limit to prove the following: $\lim_{x\to4}\frac{x-2}{\sqrt{x}+2}=\frac{1}{2}$

Use the definition of a limit to prove the following: $$\lim_{x\to4}\frac{x-2}{\sqrt{x}+2}=\frac{1}{2}$$ I'm trying to prove that: $$\forall\varepsilon>0,\exists\delta>0; \\~\\ 0<|x-4|<\delta\Longrightarrow\left|\frac{x-2}{\sqrt{x}+2}-\frac{1}{2}\right|<\varepsilon$$ I don't know what to do with $$\frac{x-2}{\sqrt{x}+2}-\frac{1}{2}$$ I've tried using the triangle inequality, but I get wrong results.

I know for now: $$\frac{1}{\sqrt{x}+2}<\frac{1}{\sqrt{3}+2}$$

Note that\begin{align}\frac{x-2}{\sqrt x+2}-\frac12&=\frac{2x-\sqrt x-6}{2\sqrt x+4}\\&=\frac{2(x-4)-\left(\sqrt x-2\right)}{2\sqrt x+4}\\&=\left(\sqrt x-2\right)\frac{2\sqrt x+3}{2\sqrt x+4}\\&=(x-4)\frac{2\sqrt x+3}{\left(\sqrt x+2\right)\left(2\sqrt x+4\right)}\\&\leqslant(x-4)\frac{2\sqrt x+3}8,\end{align}and therefore$$\left|\frac{x-2}{\sqrt x+2}-\frac12\right|\leqslant|x-4|\frac{2\sqrt x+3}8.$$So, if $$|x-4|<1$$, you have $$x<5$$, and therefore\begin{align}\frac{2\sqrt x+3}8&<\frac{2\sqrt5+3}8\\&<\frac{2\times3+3}8\\&=\frac98\\&<2.\end{align}Therefore, given $$\varepsilon>0$$, if you take $$\delta=\min\left\{1,\frac\varepsilon2\right\}$$, you have$$|x-4|<\delta\implies\left|\frac{x-2}{\sqrt x+2}-\frac12\right|<\varepsilon$$
In this specific case since the form is not undetermined (i.e. not $$\frac 00$$), I think it is easier to show separately $$|(x-2)-2|\to 0$$ and $$|(\sqrt{x}+2)-4|\to 0$$, and then that the ratio $$\to \frac 24$$, because rationalizing the denominator will not bring much simplification actually.
For the square root use $$|\sqrt{x}-2|=\frac{|x-4|}{|\sqrt{x}+2|}$$ and bound the denominator (e.g. $$|x-4|<3\implies x>1$$, etc.)
You used $$x>3$$ instead, it is not wrong, but more complicated, I'd rather have denominator $$<\frac 13$$ than $$<\frac 1{\sqrt{3}+2}$$.
In the end use or show that $$x_n\to a$$ and $$y_n\to b\neq 0$$ then $$\dfrac{x_n}{y_n}\to \dfrac{a}{b}$$.