Is $f:(0,\infty) \to \mathbb{R}$, $f(x):=x^{\sin(x)}$ uniformly continuous on $(0,1]$?

Question

Is $f:(0,\infty) \to \mathbb{R}$, $f(x):=x^{\sin(x)}$ uniformly continuous on $(0,1]$?

Hello,

i showed that $f'(x)=x^{\sin(x)}(1/x\cdot \sin(x)+\cos(x)\ln(x))$. Since $f$ is differentiable on $(0,\infty)$ it is uniformly continuous on $(0,1]$.

Is that correct?

Answer 1

No, it is not correct. By that argument, $x\mapsto\frac1x$ would also be uniformly continuous on $(0,1]$.

But the limit $\lim_{x\to0}x^{\sin x}$ exists in $\mathbb R$. It is equal to $1$, and therefore you can extend $f$ to a continuous function $F\colon[0,1]\longrightarrow\mathbb R$. Since $[0,1]$ is a closed and bounded interval, $F$ is uniformly continuous. Therefore, $f$ is uniformly continuous.