# Thread: Show that sqrt(x) is uniformly continuous on [0,infinity)

1. ## Show that sqrt(x) is uniformly continuous on [0,infinity)

Yikes! I feel like were all having trouble with uniform continuity...

Use the definition of uniform continuity to show that $\sqrt{x}$ is uniformly continuous on [0, $\infty$) (relative to [0, $\infty$))

For this, do I have to show that $\sqrt{x}$ is uniformly continuous on [0,2] and then show that it is uniformly continuous on [1, $\infty$) before I can show its uniformly continuous on [0, $\infty$)?

2. Originally Posted by tn11631
Yikes! I feel like were all having trouble with uniform continuity...

Use the definition of uniform continuity to show that $\sqrt{x}$ is uniformly continuous on [0, $\infty$) (relative to [0, $\infty$))

For this, do I have to show that $\sqrt{x}$ is uniformly continuous on [0,2] and then show that it is uniformly continuous on [1, $\infty$) before I can show its uniformly continuous on [0, $\infty$)?
That is certainly the most convenient way to do it. The usual conditions ensuring that a function is uniformly continuous are either (a) it is continuous on a bounded closed interval, or (b) it is differentiable with a bounded derivative.

The function $\sqrt{x}$ on the interval $[0,\infty)$ does not satisfy either of those conditions. But it satisfies (a) and (b) on the overlapping subintervals $[0,2]$ and $[1,\infty)$ respectively. From that, you can deduce that it is uniformly continuous on the whole interval.

3. Lemma: $|\sqrt{x}-\sqrt{y}| \le \sqrt{|x-y|}$ for any $x,y \ge 0$.

Proof:

Let $\epsilon > 0$ and choose $\delta = \epsilon^{2}$. Then, for $x,y \in [0,\infty )$, $|x-y|<\delta$ implies that $|\sqrt{x}-\sqrt{y}| \le \sqrt{|x-y|} < \sqrt{\epsilon^{2}} = \epsilon$. Thus, $f(x) = \sqrt{x}$ is uniformly continuous on $[0, \infty )$. Q.E.D.

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# prove sqrt(x) is conti

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