I need help with this proof. I have to use the definition of continuity to prove that the function f defined by f(x)=sqrt x is continuous at every nonnegative real number.
I'll leave it to you to prove that it's continuous at 0 (it's quite easy).
Suppose x is non-zero and let $\displaystyle \varepsilon >0$ be given. Then, we seek $\displaystyle \delta > 0$ such that
$\displaystyle \left|\sqrt{x}-\sqrt{y}\right| < \varepsilon .$
Use the fact that $\displaystyle \left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right) = x-y$ and the assumption that x is nonzero and you should find the desired $\displaystyle \delta$ quite easily.