Suppose that $f(x) \ge 0$ and $\lim_{x \to c} f(x) = L$. Prove $\lim_{x \to c} \sqrt{f(x)} = \sqrt{L}$ Suppose that $f(x) \ge 0$ in some deleted neighborhood of $c$, and that $\lim_{x \to c} f(x) = L$. Prove that $\lim_{x \to c} \sqrt{f(x)} = \sqrt{L}$ under the two different assumptions on $L$:
 $L=0$ and $L>0$
 A: You can do it quickly using that $x \mapsto \sqrt x$ is continuous on $[0,\infty)$, adjusting $f$ to take the value $L$ at $c$ makes it continuous, and the fact that a composition of continuous functions is continuous. To do it directly, the identity $$\sqrt a - \sqrt b = \frac{a-b}{\sqrt a + \sqrt b}$$
for $a,b>0$ is useful.
A: By hypothesis, given $\eta >0$ exists $\delta$ such that $|x-c|<\delta$ then $|f(x)-L|<\eta$. So, you have to prove that give $\epsilon>0$ exists $\lambda>0$ such that $|x-c|<\lambda$ implies $|\sqrt{f(x)}-\sqrt L|<\epsilon$. Now let us suppose $L\neq 0$ then
$$|f(x)-L|=\frac{|\sqrt{f(x)}-\sqrt L|(\sqrt{f(x)}+\sqrt L)}{\sqrt{f(x)}+\sqrt L}=
\frac{|f(x)-L|}{\sqrt{f(x)}+\sqrt L}$$
Note that $\sqrt{f(x)}+\sqrt L\geq \sqrt L$ then 
$$|f(x)-L|=\frac{|\sqrt{f(x)}-\sqrt L|(\sqrt{f(x)}+\sqrt L)}{\sqrt{f(x)}+\sqrt L}=
\frac{|f(x)-L|}{\sqrt{f(x)}+\sqrt L}\leq 
\frac{|f(x)-L|}{\sqrt L}$$
Now take $\eta=\epsilon\sqrt L$  then exists $\delta$ such that $|x-c|<\delta$ implies $|f(x)-L|<\eta$ then taking $\lambda =\delta$, and we have $|x-c|<\lambda$ implies
$$|f(x)-L|=\frac{|\sqrt{f(x)}-\sqrt L|(\sqrt{f(x)}+\sqrt L)}{\sqrt{f(x)}+\sqrt L}=
\frac{|f(x)-L|}{\sqrt{f(x)}+\sqrt L}\leq 
\frac{|f(x)-L|}{\sqrt L}<\frac {\eta} {\sqrt L}=\epsilon$$
and the result follows. Suppose now $L=0$, then ginven $\epsilon^2$ exists $\delta>0$ such that $|x-c|<\delta $ implies $|f(x)|<\epsilon^2$, since $\epsilon$ is arbitrary in the definition of limite, then $|\sqrt{f(x)}|<\epsilon$ provided $|x-c|<\delta$.
