# Finding a such that x^a*sin(1/x) is uniformly continuous.

Assuming that $\sin x$ is continuous on $\mathbb R$, find all real $\alpha$ such that $x^\alpha\sin (1/x)$ is uniformly continuous on the open interval (0,1).

I'm guessing that I need to show that $x^\alpha\sin x$ is continuously extendable to [0,1]. Doing that for $x=1$ is pretty trivial, but I am having trouble doing that for $x=0$. I believe that the $\lim_{x\to 0}x^\alpha\sin (1/x)=0$, but how can I find what $f(0)$ equals?

You're on the right track. If you can show that $f_\alpha(x) = x^\alpha \sin(1/x)$ can be continuously extended to $[0,1]$, you are done. We will show that $\lim_{x \to 0} f_\alpha(x)$ exists exactly for $\alpha > 0$ (as is 0 in this case). That means that for $\alpha > 0$, the definition $f_\alpha(0) := 0$ makes $f_\alpha\colon [0,1]\to \mathbb R$ continuous, hence uniformly (as $[0,1]$ is compact). If $\alpha > 0$, we have $$|f_\alpha(x)| \le x^\alpha \cdot |\sin(1/x)|\le x^\alpha \to 0, \quad x \to 0$$ If $\alpha \le 0$, consider $x_n = 1/(\pi/2 + n\pi)$, then $x_n \to 0$, but $$f_\alpha(x_n) = (\pi/2+ n\pi)^{|\alpha|} (-1)^n$$ and this doesn't converge for $n \to \infty$.
Hence, $f_\alpha$ is uniformly continuous on $(0,1)$ iff $\alpha > 0$.