Alternative proof that $f(x) = $ sin $\frac{1}{x^x}$ is not uniformly continuous on (0,1]?

Question

My idea is to make use of this theorem:

"A real-valued funtion $f$ on $(a,b)$ is uniformly continuous on $(a,b)$ if and only if it can be extended to a continuous function $f$ on $[a,b]$."

My reasoning is that since sin $\frac{1}{x^x}$ oscillates infinitely near the origin, it cannot be made continuous by extending $f$ by assigning any value to $f(x)$ for the case when $x = 0$.

Is this a valid line of reasoning? How could I make it mathematically precise?

Answer 1

It will not work, since$$\lim_{x\to0^+}\sin\left(\frac1{x^x}\right)=\sin\left(\lim_{x\to0^+}\frac1{x^x}\right)=\sin(1).$$Actually, it follows from this that your function is continuous on $(0,1]$, since you can extend it to a continuous function from $[0,1]$ into $\Bbb R$.