proving root of limit for root of function

Question

I'm trying to prove: if $\lim _{x\to x0}\:f\left(x\right)=\:L$ then $\lim _{x\to x0}\sqrt{f\left(x\right)}=\:\sqrt{L}$ for $L\ge 0$.

I think i've proved it well for $L=0$, saying $\varepsilon$ is $\varepsilon^2$, but i'm not sureץ Need and idea how to prove it for $L>0$.

Answer 1

Let $L>0$ . Then $|\sqrt{f(x)}-\sqrt{L}|= \frac{|f(x)-L|}{\sqrt{f(x)}+\sqrt{L}} \le \frac{|f(x)-L|}{\sqrt{L}} $

It follows that $\lim _{x\to x0}\sqrt{f\left(x\right)}=\:\sqrt{L}$