Answer
$$\lim_{x\to 4}\sqrt{x}=2 $$
Work Step by Step
Let $\epsilon\gt0$ be given. We bound $|\sqrt{x}-2|$ by multiplying $\dfrac{\sqrt{x}+2}{\sqrt{x}+2}$
\begin{align*}
|\sqrt{x}-2|&=\left|\sqrt{x}-2\left(\frac{\sqrt{x}+2}{\sqrt{x}+2}\right)\right|\\
&=\left|\frac{x-4}{\sqrt{x}+2}\right|\\
&=|x-4|\left|\frac{1}{\sqrt{x}+2}\right|
\end{align*}
assume $\delta\lt1,$ so that $|x-4|\lt1,$ and hence $\sqrt{x}+2\gt\sqrt{3}+2\gt3 .$ This gives us
\begin{align*}
|\sqrt{x}-2|&=|x-4|\left|\frac{1}{\sqrt{x}+2}\right|\\
&\lt|x-4| \frac{1}{3}
\end{align*}
Let $\delta=\min (1,3 \epsilon) .$ If $|x-4|\lt\delta$
$$
|\sqrt{x}-2|=|x-4|\left|\frac{1}{\sqrt{x}+2}\right|\lt|x-4| \frac{1}{3}\lt\delta \frac{1}{3}\lt3 \epsilon \frac{1}{3}=\epsilon
$$
Hence
$$\lim_{x\to 4}\sqrt{x}=2 $$
