How do you prove this limit: $\lim_{x \to 4} \sqrt{x} = 2$?

Question

$\lim_{x \to 4} \sqrt{x} = 2$

How would I prove this? It is an example in the textbook, but I am pretty confused.

Answer 1

Let $\epsilon > 0$. You want to find $\delta > 0$ such that $|x-4|<\delta$ implies $|\sqrt{x} - 2| < \epsilon$. This is merely the defintion of limit.

Do note that: $$|\sqrt{x}-2| = |\sqrt{x}-2|\frac{|\sqrt{x}+2|}{|\sqrt{x}+2|} = \frac{|x-4|}{|\sqrt{x}+2|} < \frac{\delta}{|\sqrt{x}+2|} \color{red}{< \epsilon}.$$We must find some condition on $|\sqrt{x}+2|$ to get that last red inequality that we want. Here this is simple, because: $$|\sqrt{x}+2| = \sqrt{x}+2 \geq 2 > 0 \implies \frac{1}{|\sqrt{x}+2|} \leq \frac{1}{2},$$so that $\delta < 2\epsilon$ will work.

You can read my answer here, in which I explain things more thoroughly.