Prove that `f(x)=sqrt(|x|-x)` is continuous for all `xgeq0.`

Answer

Function is continuous at `x ge 0 `

Explanation

\begin{aligned}
f(x) &=\sqrt{|x|-x} \quad x>, 0 \\
&f\left(0^{+}\right) \& f(0) . \\ \\
& \lim _{x \rightarrow 0^{+}} f(x) = \lim _{h \rightarrow 0} f(0+h) \\
&=\lim _{h \rightarrow 0} \sqrt{|0+h| -(0+h)}=\sqrt{|1-h| -h} \\
&=0 \\ \\
f(0)=& \sqrt{0-0} \\
&=0 \\
...