Discussion
\(f(x) = \sin x, x \in \mathbf{R}\)
Answer: C
\(\sin x\) is Lipschitz and uniformly continuous.
By the mean value theorem, \(\sin x - \sin y = \cos c\,(x - y)\) for some \(c\) between \(x\) and \(y\). Since \(|\cos c| \le 1\),
\[ |\sin x - \sin y| \le |x - y| \]
so sine is Lipschitz with constant 1. Every Lipschitz function is uniformly continuous: given \(\varepsilon \gt 0\), \(\delta = \varepsilon\) works for every pair of points at once.
So the options denying uniform continuity are wrong, and 'uniformly continuous but not Lipschitz' fails because the Lipschitz bound holds. That last description fits a function such as \(\sqrt{x}\) on [0, 1], not \(\sin x\).
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