Part BCSIR NET December 2024one-subsequence-bounds-the-liminf-only-one-way

One subsequence bounds the liminf only one way

Let $(x_{n})_{n} \geq_{1}$ be a sequence of real numbers that has a decreasing subsequence $(x_{(} n_{k}))_{k} \geq_{1}$ . Assume that $lim (k \to \infty) x_{(} n_{k}) = 2025$ . Which of the following statements is necessarily true?

A. $lim in f (n \to \infty) x_{n} \geq 2025$
B. $lim sup (n \to \infty) x_{n} \leq 2025$
C. $lim in f (n \to \infty) x_{n} \leq 2025$ ✓
D. $lim in f (k \to \infty) x_{(} n_{k}) > 2025$

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

Related counterexample: limsup(aₙ + bₙ) = limsup aₙ + limsup bₙ

One subsequence bounds the liminf only one way

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