1. ## Continuity of f(x)

Being new to continuity, how do I use the definition of continuity to show that

f(x) = sqrt(x) is continuous on the domain of f?

Any help would be very appreciated!

3. ## Re: Continuity of f(x)

Recap the definition of continuity: The function $f:I\to\mathbb R$ is said to be continuous at the point $x_0\in I$ iff for all $\varepsilon>0,$ there exists $\delta>0$ such that for all $x\in I,$ $\left|x-x_0\right|<\delta\Rightarrow\left|f(x)-f(x_0)\right|<\varepsilon,$ and $f$ is said to be continuous on $I$ iff it is continuous at every point in $I.$

For $f(x)=\sqrt x,$ $I=\left\{x\in\mathbb R:x\geqslant0\right\}.$

First show that $f(x)=\sqrt x,$ is continuous at $x_0=0.$ Given $\varepsilon>0,$ take $\delta=\varepsilon^2.$ Then for all $x\geqslant0,$ $|x-0|<\delta$ $\implies$ $|x|<\varepsilon^2$ $\implies$ $\left|\sqrt x\right|=\left|\sqrt x-\sqrt0\right|<\varepsilon.$

Now show that $f$ is continuous at $x_0>0.$ Given $\varepsilon>0,$ take $\delta=\varepsilon\sqrt x_0.$ Then $\left|x-x_0\right|<\delta$ $\implies$ $\left|\sqrt x-\sqrt{x_0}\right|\left|\sqrt x+\sqrt{x_0}\right|<\delta$ $\implies$ $\left|\sqrt x-\sqrt x_0\right|<\frac{\delta}{x+\sqrt{x_0}} \leqslant\frac{\delta}{\sqrt{x_0}}=\varepsilon.$