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Let \((X_{n})\) be a sequence of real numbers and \(x\in\mathbf{R}\). Then which of the following is true ?

This question was previously asked in
MH SET 2020 Paper 2 Philosophy Question Paper (27-Dec-2020)
The correct answer is

If \(\lim\limits_{n\to\infty} x_{n}\lt x\), then \(x_{n}\lt x\) for infinitely many numbers \(n\)

Let \(L=\lim\limits_{n\to\infty} x_{n}\) with \(L\lt x\). Take \(\varepsilon=\frac{x-L}{2}\gt 0\); by the definition of the limit there is \(N\) with \(|x_{n}-L|\lt\varepsilon\) for all \(n\geq N\).

For those \(n\), \(x_{n}\lt L+\varepsilon=\frac{L+x}{2}\lt x\). So \(x_{n}\lt x\) for every \(n\geq N\), which is certainly infinitely many indices.

A limit only controls the tail, not every term: \(x_{n}=\frac{1}{n}\) has limit \(0\) and \(0\leq 0\), but \(x_{1}=1\gt 0\), so the claim for all \(n\) fails.

A different limit does not force terms on both sides of \(x\): the constant sequence \(x_{n}=1\) with \(x=0\) has limit \(1\neq 0\), yet no term is below \(0\).

A divergent sequence can still have convergent subsequences: \(x_{n}=(-1)^{n}\) has no limit, but \(x_{2n}=1\) converges to \(x=1\).

Check: for \(x_{n}=\frac{1}{n}\) and \(x=\frac{1}{2}\), the limit \(0\) is below \(x\), and indeed \(x_{n}\lt\frac{1}{2}\) for every \(n\geq 3\), infinitely many indices.

Hence, If \(\lim\limits_{n\to\infty} x_{n}\lt x\), then \(x_{n}\lt x\) for infinitely many numbers \(n\).

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