# Thread: Is the sqrt(x) uniformly continuous on [0,inf)?

1. ## Is the sqrt(x) uniformly continuous on [0,inf)?

Is $\sqrt{x}$ uniformly continuous on $[0,\infty)$? Justify your answer.

(Note: It's slope at 0 is infinite)

2. yes, it is.

let's break that set into $[0,1]\cup[1,\infty).$

note that the first set is compact, so $f$ is automatically uniformly continuous there, but as for the second set, for each $x,y\ge1$ is $\left| \sqrt{x}-\sqrt{y} \right|=\frac{\left| x-y \right|}{\sqrt{x}+\sqrt{y}}\le \frac{1}{2}\left| x-y \right|,$ and this satisfies Lipschitz condition, hence we have uniform continuity on the second set, and finally on the whole set.

3. Originally Posted by Krizalid
yes, it is.

let's break that set into $[0,1]\cup[1,\infty).$

note that the first set is compact, so $f$ is automatically uniformly continuous there, but as for the second set, for each $x,y\ge1$ is $\left| \sqrt{x}-\sqrt{y} \right|=\frac{\left| x-y \right|}{\sqrt{x}+\sqrt{y}}\le \frac{1}{2}\left| x-y \right|,$ and this satisfies Lipschitz condition, hence we have uniform continuity on the second set, and finally on the whole set.
You could also note that it's Lipschitz continuous on [1, infinity) since its derivative is bounded there.