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

• Oct 25th 2009, 06:39 PM
cgiulz
Is the sqrt(x) uniformly continuous on [0,inf)?
Is $\displaystyle \sqrt{x}$ uniformly continuous on $\displaystyle [0,\infty)$? Justify your answer.

(Note: It's slope at 0 is infinite)
• Oct 25th 2009, 06:56 PM
pickslides
• Oct 25th 2009, 07:00 PM
Krizalid
yes, it is.

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

note that the first set is compact, so $\displaystyle f$ is automatically uniformly continuous there, but as for the second set, for each $\displaystyle x,y\ge1$ is $\displaystyle \left| \sqrt{x}-\sqrt{y} \right|=\frac{\left| x-y \right|}{\sqrt{x}+\sqrt{y}}\le \frac{1}{2}\left| x-y \right|,$ and this satisfies Lipschitz condition, hence we have uniform continuity on the second set, and finally on the whole set.
• Oct 25th 2009, 07:09 PM
rn443
Quote:

Originally Posted by Krizalid
yes, it is.

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

note that the first set is compact, so $\displaystyle f$ is automatically uniformly continuous there, but as for the second set, for each $\displaystyle x,y\ge1$ is $\displaystyle \left| \sqrt{x}-\sqrt{y} \right|=\frac{\left| x-y \right|}{\sqrt{x}+\sqrt{y}}\le \frac{1}{2}\left| x-y \right|,$ 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.