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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 10, Problema 10.2.19

13–52. Limits of sequences
Find the limit of the following sequences or determine that the sequence diverges.


{1 + cos(1⁄n)}

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Identify the given sequence: \(a_n = 1 + \cos\left(\frac{1}{n}\right)\), where \(n\) is a positive integer and \(n \to \infty\).
Recall that as \(n\) approaches infinity, the term \(\frac{1}{n}\) approaches 0, so we need to analyze the behavior of \(\cos\left(\frac{1}{n}\right)\) as its argument approaches 0.
Use the fact that \(\cos(x)\) is continuous and \(\cos(0) = 1\), so \(\lim_{x \to 0} \cos(x) = 1\). Therefore, \(\lim_{n \to \infty} \cos\left(\frac{1}{n}\right) = 1\).
Apply the limit to the entire sequence: \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(1 + \cos\left(\frac{1}{n}\right)\right) = 1 + \lim_{n \to \infty} \cos\left(\frac{1}{n}\right)\).
Combine the results to conclude that the limit of the sequence is \(1 + 1 = 2\).

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Concetti chiave

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Limit of a Sequence

The limit of a sequence is the value that the terms of the sequence approach as the index goes to infinity. If the terms get arbitrarily close to a specific number, the sequence converges to that limit; otherwise, it diverges.
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Introduction to Sequences

Behavior of the Cosine Function Near Zero

The cosine function is continuous and approaches 1 as its argument approaches 0. Understanding that cos(1/n) approaches cos(0) = 1 as n → ∞ is key to evaluating the limit of sequences involving cosine of reciprocal terms.
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Graph of Sine and Cosine Function

Substitution and Limit Laws

Limit laws allow the evaluation of limits by substituting the limit of the inner function into the outer function when the outer function is continuous. Here, since cosine is continuous, we can find the limit by substituting the limit of 1/n into cos(1/n).
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Finding Limits by Direct Substitution