Exercise · ZW-SEQUENTIAL-CONTINUITY-007 · ★★☆ Intermediate
Let \(f(x)=\sin(1/x)\) for \(x\ne0\) and \(f(0)=0\). With \(x_n=\dfrac{1}{\pi/2+2\pi n}\), what does \(f(x_n)\) converge to?
Choose one option.
Show answer
- C. \(1\)
Exercise · ZW-SEQUENTIAL-CONTINUITY-007 · ★★☆ Intermediate
Choose one option.
Here \(1/x_n=\pi/2+2\pi n\) and the sine of that equals \(1\). Since \(x_n\to0\) but \(f(x_n)=1\ne f(0)\), it is not continuous at \(0\).
A function \(f\) is sequentially continuous at \(c\) if, for every sequence \((x_n)\) in the domain converging to \(c\), the sequence of images \(f(x_n)\) converges to \(f(c)\).
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