Ch. 11 - Power Series

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 11 - Power Series

Problema 11.4.69

Capitolo 11, Problema 11.4.69

A limit by Taylor series Use Taylor series to evaluate lim ₓ→₀ ((sin x)/x)¹/ˣ²

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Recognize that the limit is of the form \(\lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{1}{x^2}}\). Since direct substitution leads to an indeterminate form, we use the Taylor series expansion to simplify the expression inside the limit.

Recall the Taylor series expansion of \(\sin x\) around \(x=0\): \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\). Substitute this into the expression \(\frac{\sin x}{x}\) to get \(\frac{\sin x}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \cdots\).

Rewrite the original limit as \(\lim_{x \to 0} \left( 1 - \frac{x^2}{6} + \cdots \right)^{\frac{1}{x^2}}\). This is a limit of the form \(\left(1 + f(x)\right)^{g(x)}\) where \(f(x) \to 0\) and \(g(x) \to \infty\) as \(x \to 0\).

Use the fact that \(\lim_{x \to 0} (1 + f(x))^{g(x)} = e^{\lim_{x \to 0} f(x) g(x)}\) if the latter limit exists. Here, identify \(f(x) = -\frac{x^2}{6} + \cdots\) and \(g(x) = \frac{1}{x^2}\).

Calculate the exponent limit: \(\lim_{x \to 0} f(x) g(x) = \lim_{x \to 0} \left(-\frac{x^2}{6} + \cdots \right) \cdot \frac{1}{x^2}\). Simplify this expression to find the exponent of \(e\), which will give the value of the original limit.

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Taylor Series Expansion

A Taylor series represents a function as an infinite sum of terms calculated from the function's derivatives at a single point. For functions like sin(x), the series expansion around x=0 helps approximate values near zero, making it easier to analyze limits involving complex expressions.

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Limits Involving Indeterminate Forms

Limits that result in forms like 0^∞ or 1^∞ are indeterminate and require special techniques to evaluate. Using series expansions or logarithmic transformations can simplify these expressions, allowing the limit to be computed accurately.

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When limits involve expressions raised to powers that approach zero or infinity, rewriting the expression using logarithms and exponentials helps. Taking the natural logarithm transforms the power into a product, simplifying the limit evaluation before exponentiating back to find the original limit.

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