Uniform Continuity on unbounded interval

Question

I have trouble in the following problems:

1. Let $f:(0,\infty)\to\mathbb{R}$ be a function defined by $f(x)=x^{\alpha}\sin(x^{\alpha})$, where $\alpha\in\mathbb{R}$.

   Find the range of $\alpha$ so that $f$ is uniform continuous on $(0,\infty)$.

2. Let $g:(0,\infty)\to\mathbb{R}$ be a function defined by $g(x)=x^{\beta}\cos(x^{\beta})$, where $\beta\in\mathbb{R}$.

   Find the range of $\beta$ so that $f$ is uniform continuous on $(0,\infty)$.

Using the Continuosly Extension Theorem and Bounded Derivative property, i guess $0<\alpha,\beta\le\frac{1}{2}$.

But, i can't prove the other case that $f$ is not uniformly continuous unless $0<\alpha,\beta\le\frac{1}{2}$.

Give some hint or advice. Thank you.

Answer 1

Hint 1:

For $\alpha > 1/2$, take $x_n = (n\pi + 1/n)^{1/\alpha}$ and $y_n = (n\pi)^{1/\alpha}$ and show that as $n \to \infty$ we have $|x_n-y_n| \to 0$ but $|f(x_n) - f(y_n)| \not\to 0$.

Hint 2:

Using the binomial expansion

$$|x_n - y_n| = (n\pi)^{1/\alpha}\left[\left(1 + \frac{1}{n^2\pi}\right)^{1/\alpha} - 1\right]= (n\pi)^{1/\alpha}\left[1 + \frac{1}{\alpha n^2\pi } + O(n^{-4})-1\right]$$