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

One subsequence bounds the liminf only one way

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

A. $\lim \inf(n\to\infty) x_{n} \ge 2025$
B. $\lim \sup(n\to\infty) x_{n} \le 2025$
C. $\lim \inf(n\to\infty) x_{n} \le 2025$ ✓ Correct
D. $\lim \inf(k\to\infty) x_(n_{k}) > 2025$

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, and why this one tests limit assumed to exist.

See pricing

50 are analysed free — try those first.

The trap it tests

Limit assumed to exist

Reasoning with a limit before establishing there is one. limsup is not lim.

Drill statements like this

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

One subsequence bounds the liminf only one way

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