Prove this limit with Epsilon-Delta Definiton

Question

$$\lim_{x\to x_0}\sqrt x=\sqrt{x_0}$$

Hello guys, İ have tried to prove this limit but i couldn't.

Answer 1

Since$$\sqrt x-\sqrt{x_0}=\frac{\bigl(\sqrt x-\sqrt{x_0}\bigr)\bigl(\sqrt x+\sqrt{x_0}\bigr)}{\sqrt x+\sqrt{x_0}}=\frac{x-x_0}{\sqrt x+\sqrt{x_0}}<\frac{x-x_0}{\sqrt{x_0}},$$you can take $\delta=\sqrt{x_0}\varepsilon$. Of course, I am assuming that $x_0\neq 0$. Can you do it when $x_0=0$?

Answer 2

You need to establish a relation between $|x-x_0|$ and $|\sqrt x-\sqrt{x_0}|$. Hopefully you will see that $|x-x_0|=|\sqrt x-\sqrt{x_0}|\cdot|\sqrt x+\sqrt{x_0}|$, or

$$|\sqrt x-\sqrt{x_0}|=\frac{|x-x_0|}{|\sqrt x+\sqrt{x_0}|}.$$

Then if $x$ remains close enough to $x_0$, say $x\ge\dfrac{x_0}2$, you can bound the denominator.