Exercise · ZW-SEQUENTIAL-CONTINUITY-005 · ★☆☆ Basic

Let \(f(x)=\dfrac{x^2-1}{x-1}\) if \(x\ne1\) and \(f(1)=3\). With \(x_n=1+1/n\), what does \(f(x_n)\) converge to?

Choose one option.

Show answer
  • D. \(2\)

Explanation

For \(x\ne1\), \(f(x)=x+1\), so \(f(x_n)=2+1/n\to2\ne f(1)=3\): there is a discontinuity at \(1\).

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Review the concept: Sequential continuity

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 \(f(c)\).

\[x_n\to c\ \Longrightarrow\ f(x_n)\to f(c)\]

More practice problems: sequential continuity

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