# Why is f(x)=sqrtx continuous?

For every number $a$ in $\left(0 , \infty\right)$, we have ${\lim}_{x \rightarrow a} \sqrt{x} = \sqrt{a}$. That is the definition of continuous at $a$.
At $0$, we have ${\lim}_{x \rightarrow {0}^{+}} \sqrt{x} = \sqrt{0}$ which is the definition of continuous from the right at $0$.
So $\sqrt{x}$ is continuous on $\left[0 , \infty\right)$.
${\lim}_{x \rightarrow a} \sqrt{x} = \sqrt{a}$ can be proven using the definition of limit.