We need to find which are uniform continuous (UC) on a) $(0,1)$ and b) $(0,\infty)$. I have done, could you confirm me, if I am wrong any where?
$\frac{1}{(1-x)}$
$\frac{1}{(2-x)}$
$\sin x$
$\sin(1/x)$
$x^{1/2}$
$x^3$
1) is not UC on a) because limit does not exist when $x\rightarrow 1$, on b) it is discontinous (disco) at $x=1$
2) is UC on a) and it is disco at $x=2$ so not UC on b)
3) is UC on a) and also on b) as we can show by the inequality $|\sin x-\sin y|<|x-y|$
4) is not UC on a) as limit does not exist, and also not UC on b) I am not clear enough for this one.
5) is UC on a) and not UC on b) as derivative is not bounded near $0$
6) is UC on a) and not UC on b) as derivative is not bounded.