Exercise · ZW-UNIFORM-CONTINUITY-007 · ★★☆ Intermediate
To prove \(\sin x\) is uniformly continuous on \(\mathbb R\) one uses \(|\sin x-\sin y|\le|x-y|\). Which \(\delta\) works?
Choose one option.
Show answer
- A. \(\delta=\varepsilon\)
Explanation
If \(|x-y|<\varepsilon\) then \(|\sin x-\sin y|\le|x-y|<\varepsilon\), with \(\delta=\varepsilon\) independent of the point. \(\varepsilon/|x|\) depends on \(x\), and \(\delta=1\) fails for \(\varepsilon<1\).
Review the concept: Uniform continuity
A function \(f\) is uniformly continuous on a set \(A\) if for each \(\varepsilon>0\) there is a \(\delta>0\), valid for all points at once, that controls \(|f(x)-f(y)|\) by \(|x-y|\) in \(A\).
\[\forall\varepsilon>0\ \exists\delta>0:\ x,y\in A,\ |x-y|<\delta\Rightarrow|f(x)-f(y)|<\varepsilon\]
More practice problems: uniform continuity
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