i got stuck on this problem
How can prove it?
i dont wanna use that sin(1/x)=1/x-1/(6$x^3$)+...
2025-01-13 02:18:24.1736734704
Is $x^3\sin\left(\frac{1}{x}\right)$ is uniformly continuous in (0,$\infty$)?
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Here is a hint to get you started: as $x \to \infty$, $\sin \frac 1x$ is approximately equal to $\frac 1x$, so for large $x$, $x^3 \sin \frac 1x$ is approximately equal to $x^2$. Is $x^2$ uniformly continuous on $(0,\infty)$?