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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 10.2.11

Compare the growth rates of {n¹⁰⁰} and {eⁿ⁄¹⁰⁰} as n → ∞.

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1
Identify the two functions to compare: \(f(n) = n^{100}\) and \(g(n) = e^{\frac{n}{100}}\).
Recall that to compare growth rates as \(n \to \infty\), we often consider the limit of their ratio, such as \(\lim_{n \to \infty} \frac{f(n)}{g(n)}\) or \(\lim_{n \to \infty} \frac{g(n)}{f(n)}\).
Set up the limit \(L = \lim_{n \to \infty} \frac{n^{100}}{e^{\frac{n}{100}}}\) to analyze which function grows faster.
Apply L'Hôpital's Rule if the limit is an indeterminate form like \(\frac{\infty}{\infty}\) by differentiating numerator and denominator with respect to \(n\) repeatedly, or use properties of exponential and polynomial functions to reason about the limit.
Conclude the comparison based on the limit: if \(L = 0\), then \(g(n)\) grows faster; if \(L = \infty\), then \(f(n)\) grows faster; if \(L\) is a finite nonzero constant, they grow at comparable rates.

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Polynomial growth involves functions where the variable is raised to a fixed power, like n^100. Although these functions grow quickly for large n, their growth rate is slower compared to exponential functions when n approaches infinity.
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To compare growth rates as n → ∞, we analyze the limit of the ratio of the two functions. If the limit is zero, the numerator grows slower; if infinite, it grows faster. This method helps determine which function dominates in the long run.
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