TITLE: Continuous function proof by definition
QUESTION [9 upvotes]: Prove that if $f$ is defined for $x\ge 0$ by $f(x)=\sqrt x$, then $f$
  is continuous at every point of its domain.

Definition of a continuous function is:
Let $A\subseteq\mathbb{R}$ and let $f:A\to\mathbb{R}$. Denote $c\in A$.
Then $f(x)$ is continuous at $c$ iff for every $\varepsilon>0$, $\exists$ $\delta>0$ such that 
$|x-c|<\delta\implies |f(x)-f(c)|<\varepsilon.$
My attempt:
We know that the function $f: x\to \mathbb{R}$, where $x\in [0,\infty)$ is defined to be $f(x)=\sqrt x$. So, for $0\le x<\infty$, then 
$|f(x)-f(c)|=|\sqrt x - f(c)|$ and I can't continue since I don't necessarily know what $c$ is in this case.

REPLY [2 votes]: user62089's answer is not quite right. At $c \neq 0$, user62089's work proves that $\delta = \varepsilon \sqrt{c}$ is sufficient. At $c=0$ you have a slightly different situation: you want $\delta>0$ so that $\sqrt{x}<\varepsilon$ for $x<\delta$. $\delta=\varepsilon^2$ works for this purpose.