Let $f(x) = x^2 \sin(x)$ and let $x_n=1/n$. What is the limit of the sequence $\{f(x_n)\}$?

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We are given f(x) = x^2 sin(x) and the sequence x_n = 1/n, and we are asked to find the limit of f(x_n).

First, observe that as n → ∞, x_n → 0 since x_n = 1/n. So we are effectively evaluating the behavior of f near x = 0 along the particular sequence x_n.

Use the standard bound sin(t) ≤ |t| for all real t. Applying this with t = x_n, we get |sin(x_n)| ≤ |x_n| = 1/n.

Compute the absolute value of f(x_n): |f(x_n)| = |x_n^2 sin(x_n)| ≤ x_n^2 · |sin(x_n)| ≤ x_n^2 · |x_n| = x_n^3 = (1/n)^3 = 1/n^3.

Since 1/n^3 → 0 as n → ∞, by the squeeze (or comparison) principle the sequence f(x_n) must converge to 0 as n grows large.

An alternate viewpoint is to note the standard limit sin(x) ~ x as x → 0, so f(x) ~ x^3 near 0; substituting x = x_n = 1/n gives f(x_n) ~ (1/n)^3 → 0. This corroborates the bound-based conclusion.

Because the question presents a short-answer format without explicit multiple choices, the reasoning above uses a direct bound and a basic limit argument to determine the limit of the sequence.

MTH1030 -1035 - S1 2025 MTH1030/5 Week 8 lesson quiz: Infinite sequences and series

Let \(f(x) = x^2 \sin(x)\) and let \(x_n=1/n\). What is the limit of the sequence \(\{f(x_n)\}\)?

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