Is the function $f(x)$ uniformly continuous on $(0, 1)$ if
(a) $f(x) = x\sin(x^{-2})$
(b) $f(x) = \sin(x^{-2})$
I have been using big O notation to try to solve this - since $$\sin(u) = u- {u^{3}\over3!}+{u^{5}\over5!} - \dots$$ which can be written as $$\sin(u) = u- {u^{3}\over3!}+O(u^{5}).$$
Then substituting $\displaystyle u = {1\over x^{2}}$, get $$\sin\bigg({1\over x^{2}}\bigg) = {1\over x^{2}}- {1\over(3!)x^{6}}+O\bigg({1\over x^{10}}\bigg)$$
and $$x\sin(x) = {1\over x}- {1\over(3!)x^{5}}+O\bigg({1\over x^{9}}\bigg)$$
(Am I allowed to cancel like that in big O notation?).
Where I'm stuck is how to arrange this to prove uniform continuity. I know the definition of uniform continuity, for ever $\epsilon >0$, there exists a $\delta >0$ such that $|x-c|<\delta$ and $|f(x)-f(c)|<\epsilon$, but I don't know how to rearrange the big O notation so that it fits this definition.
Since $f(x) = x\sin(1/x^2)$ has limit $0$ as $x \to 0$, we can extend $f$ to a continuous function $g : [0,1] \to \mathbb{R}$ defined by $g(x) = x\sin(1/x)^2$ on $(0,1)$ and $g(0) = 0$. Since $[0,1]$ is compact, $g$ is uniformly continuous on $[0,1]$. Therefore $g$ is uniformly continuous on $(0,1)$, hence so is $f$.
Now for the other one: Let $\delta > 0$. Choose $n \in \mathbb{N}$ with $n > 1/\delta$. Then
\begin{equation*} \begin{aligned} \left\vert \frac{1}{\sqrt{\pi n}} - \frac{1}{\sqrt{\pi(n + 1/2)}} \right\vert &= \left\vert \frac{\sqrt{n + 1/2} - \sqrt{n}}{\sqrt{\pi n(n + 1/2)}} \right\vert \\ &< \left\vert \frac{1}{\sqrt{\pi n(n + 1/2)}}\right\vert \\ &< \left\vert \frac{1}{n} \right\vert \\ &= \delta \end{aligned} \end{equation*}
But
$$\left\vert \sin\left(\frac{1}{(1/\sqrt{\pi n})^2}\right) - \sin\left(\frac{1}{(1/\sqrt{\pi(n + 1/2)})^2}\right)\right\vert = 1.$$
So $f(x) = \sin(1/x^2)$ is not uniformly continuous on $(0,1)$.