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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 10.2.23

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


{(√(4n⁴ + 3n))⁄(8n² + 1)}  

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Identify the given sequence: \(a_n = \frac{\sqrt{4n^4 + 3n}}{8n^2 + 1}\).
To find the limit as \(n \to \infty\), first analyze the dominant terms in the numerator and denominator. The highest power of \(n\) inside the square root is \(n^4\), and in the denominator it is \(n^2\).
Rewrite the numerator by factoring out \(n^4\) inside the square root: \(\sqrt{4n^4 + 3n} = \sqrt{n^4(4 + \frac{3}{n^3})} = n^2 \sqrt{4 + \frac{3}{n^3}}\).
Rewrite the denominator as \(8n^2 + 1 = n^2(8 + \frac{1}{n^2})\).
Express the sequence as \(a_n = \frac{n^2 \sqrt{4 + \frac{3}{n^3}}}{n^2 (8 + \frac{1}{n^2})} = \frac{\sqrt{4 + \frac{3}{n^3}}}{8 + \frac{1}{n^2}}\). Then, take the limit as \(n \to \infty\) by evaluating the limits of the numerator and denominator separately.

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