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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 11, Problema 11.4.14

Limits Evaluate the following limits using Taylor series.
lim ₓ→∞ x sin(1/x)

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Recognize that the limit involves the expression \(x \sin\left(\frac{1}{x}\right)\) as \(x\) approaches infinity, which suggests using the Taylor series expansion of \(\sin z\) around \(z=0\) where \(z = \frac{1}{x}\).
Recall the Taylor series expansion for \(\sin z\) around \(z=0\): \(\sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots\)
Substitute \(z = \frac{1}{x}\) into the series: \(\sin\left(\frac{1}{x}\right) = \frac{1}{x} - \frac{1}{6x^3} + \frac{1}{120x^5} - \cdots\)
Multiply the entire series by \(x\): \(x \sin\left(\frac{1}{x}\right) = x \left( \frac{1}{x} - \frac{1}{6x^3} + \frac{1}{120x^5} - \cdots \right) = 1 - \frac{1}{6x^2} + \frac{1}{120x^4} - \cdots\)
Evaluate the limit as \(x \to \infty\) by observing that all terms with \(x\) in the denominator approach zero, so the limit is the constant term remaining in the expression.

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