Any help would be appreciated.
2025-01-13 02:12:06.1736734326
Is $x^3\sin(1/x^3)$ a uniform continuous function on $(0,1)$.
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Yes, the function is uniformly continuous, because it can be continuously extended to a function on the closed interval $[0,1]$, since $$ \lim_{x\to0}x^3\sin\frac{1}{x^3}=0 $$ The function on $[0,1]$ $$ f(x)=\begin{cases} 0 & \text{if $x=0$}\\[6px] x^3\sin\dfrac{1}{x^3} & \text{if $0<x\le 1$} \end{cases} $$ is continuous on a closed and bounded interval, so it is uniformly continuous and hence any restriction thereof is also uniformly continuous.
Note that the derivative of your function is $$ 3x^2\sin\frac{1}{x^3}-\frac{3}{x}\cos\frac{1}{x^3} $$ so the derivative is not bounded on $(0,1)$. However, boundedness of the derivative is a sufficient condition for a function to be Lipschitzian, which in turn is stronger than uniform continuity.