In my lecture notes I saw the proof of $f(x) = \sqrt x$ is uniformly continous on $[0,+∞)$.
The proof goes as follows:
Given $\epsilon >0$, we pick $\delta = \epsilon ^2$. We note that $|\sqrt x-\sqrt y| ≤ |\sqrt x+\sqrt y|$.
Hence if
$|x − y| < \delta = \epsilon ^2$, then we have
$$|\sqrt x-\sqrt y|^2≤ |\sqrt x-\sqrt y||\sqrt x+\sqrt y|≤|x-y| < \epsilon ^2$$
Hence $|\sqrt x-\sqrt y| <\epsilon$ and this shows that $f(x) = \sqrt x$ is uniformly continuous on $[0,+∞)$.
Just out of curiosity, can I generalize this to show that the $n$th root (where $n>0$) of $x$ is uniformly continuous on $[0,+∞)$?
Best Answer
The trick: using the identity $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}).$$ (What is $a$ and $b$ in your case?)
EDIT: particular case $n=3$:
$$a^n-b^n=(a-b)(a^2+ab+b^2)$$
$$a-b=\frac{a^n-b^n}{a^2+ab+b^2}$$ $$ \root3\of x-\root3\of y= \frac{x-y}{(\root3\of x)^2+\root3\of x\root3\of y+(\root3\of y)^2} $$ $$\dots$$