$\sin \frac{1}{x}$ is not uniformly continuous on $[1, \infty]$
I was trying to come up with a counter example using sequence, but I could not find one.
2025-01-13 02:39:30.1736735970
$\sin \frac{1}{x}$ is not uniformly continuous on $[1, \infty]$
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If $1 \leq x < y$, then by mean-value theorem we have $$ |\sin \frac{1}{x} - \sin \frac{1}{y}| \leq |x-y|\sup_{x \leq t \leq y}|\frac{\cos (1/t)}{t^{2}}| \leq |x-y|; $$ hence the map $x \mapsto \sin (1/x): [1, \infty[ \to \mathbb{R}$ is uniformly continuous.