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I am working on the following problem:

Is $g(x)=\sin(\frac{1}{x+1})$ uniformly continuous on $(0,\infty)$?

I have two answers to this:

  1. Not Uniformly Continuous:

    In class, we learned that one way to show a function is not uniformly continuous is to find two sequences, show they get arbitrarily close -- but their images stay away from one another. So consider sequences $(x_k) = \frac{1}{-\pi/2 + 2k\pi} - 1$ , $(y_k) = \frac{1}{\pi/2 + 2k\pi} - 1$. Then $$|x_k - y_k| = \frac{1}{4k^2\pi - \pi/4}$$ Thus the sequences get arbitrarily close. The images however stay apart: $|f(x_k)-f(y_k)|=2$.

  2. Uniform Continuity

    Chain rule shows $|g'(x)| \leq 1$, thus by Mean Value Theorem $|g(a)-g(b)| \leq |a-b| \forall a,b \in (0,\infty)$.

They both seem right to me. Help!

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Your $x_k$, $y_k$ are not in $(0,\infty)$ for large $k$. Only finite number of them is positive, so you cannot think as in 1.

Intuition: $\sin(1/x)$ is uniformly continuous on $(1,\infty)$.

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