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Problem: Prove or disprove that the function $ f(x) := m x^{m} \cdot \sin \left( \dfrac{1}{x^{n}} \right) $ defined on $ [a,\infty) $, where $ a > 0 $ and $ m > n > 0 $, is uniformly continuous on $ [a,\infty) $.

Clearly, $ f $ is uniformly continuous on the bounded closed interval $ [a,b] $ for any $ b > a $, but the question is about the uniform continuity of $ f $ on the unbounded interval $ [a,\infty) $.

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Hint:

  • $\frac{y}{2} \lt \sin(y) \lt y$ for $0 \lt y \lt 1$

  • let $y = \dfrac{1}{x^n}$

  • remember $m \gt n$

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