So I was given this question on a worksheet and I'm pretty sure I have it but the solution seems a little too simple. Can someone let me know if I'm missing something?
"Why do you know, without any computation, that $f(x) = \sqrt x$ is uniformly continuous on the interval $[0.01,100]$. Then, find a $\delta$ that corresponds to an arbitrary choice of $\epsilon \gt 0$ that satisfies the definition of uniform continuity."
I think the answer is that since $[0.01,100]$ is a compact interval it is then uniformly continuous. Then for part 2 of the question I'm thinking that $\delta = \epsilon$ which seems a little trivial. Can someone help me confirm this?
You are correct on part. Since $f(x)=\sqrt{x}$ is continuous on $[0.01,100]$, and $[0.01,100]$ is compact, $f$ is necessarily uniformly continuous on $[0.01,100]$. As for the $\epsilon$ we have that following. Assume $x\ge y$, then we have $$\sqrt{|x-y|}\ge |\sqrt{x}-\sqrt{y}|$$ because by squaring both sides we see this is equivalent to $$x-y\ge x+y-2\sqrt{xy}\Leftrightarrow \sqrt{xy}\ge y$$ which is clearly true by $x\ge y$. Hence we see that one such valid $\epsilon$ is $\epsilon=\sqrt{\delta}$. By example we can see that $\epsilon=\delta$ does not work because consider for example $x=0.01$ and $x=0.11$. Then $\delta=.1$ but $|\sqrt{0.01}-\sqrt{0.11}|=0.231...>\delta$.