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For each positive integer \(n\), define \(x_n = (-1)^{n}\).

Which ONE of the following statements about the sequence \((x_n)\) is FALSE?

  • A.There exists \(\epsilon \gt 0\) such that \(|x_n - 1| \lt \epsilon\) for all positive integers \(n\).
  • B.There exists \(\epsilon \gt 0\) such that for all \(M \gt 0\) there exists a positive integer \(N \gt M\) for which \(|x_N - 1| \gt \epsilon\).
  • C.For all \(\epsilon \gt 0\) and \(M \gt 0\) there exists a positive integer \(N\) such that \(N \gt M\) and \(|x_N - 1| \lt \epsilon\).
  • D.For all \(\epsilon \gt 0\) and \(M \gt 0\) there exists a positive integer \(N\) such that \(N \gt M\) and \(|x_N - 1| \gt \epsilon\).

Answer: D

The FALSE statement is the last one: for all \(\epsilon \gt 0\) and \(M \gt 0\) there is \(N \gt M\) with \(|x_N - 1| \gt \epsilon\).

Since \(x_n = (-1)^{n}\), the distance \(|x_n - 1|\) is 0 for even \(n\) and 2 for odd \(n\). It never exceeds 2, so for \(\epsilon = 2\) (or anything larger) no \(N\) satisfies \(|x_N - 1| \gt \epsilon\). A statement claimed for all \(\epsilon\) fails.

The other three are true. The first holds with \(\epsilon = 3\), since every distance is at most 2. The second holds with \(\epsilon = 1\): beyond any \(M\) there is an odd \(N\), where the distance is \(2 \gt 1\). The third holds because beyond any \(M\) there is an even \(N\), where the distance is \(0 \lt \epsilon\). Together, the second and third say that 1 is a limit point of the sequence but not its limit.

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