Textbook Question
Remainders Find the remainder Rₙ for the nth−order Taylor polynomial centered at a for the given functions. Express the result for a general value of n.
f(x) = 1/(1 - x), a=0
Ch. 11 - Power Series
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 11, Problem 11.4.14
Limits Evaluate the following limits using Taylor series.
lim ₓ→∞ x sin(1/x)
Verified step by step guidance1
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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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Limits at Infinity
Limits at infinity describe the behavior of a function as the input grows without bound. Understanding how functions behave as x approaches infinity helps determine if the function approaches a finite value, infinity, or does not exist.
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Cases Where Limits Do Not Exist
Taylor Series Expansion
A Taylor series represents a function as an infinite sum of terms calculated from its derivatives at a single point. It approximates functions near that point, allowing complex expressions like sin(1/x) to be expanded into simpler polynomial terms for limit evaluation.
Recommended video:
08:42Taylor Series
Asymptotic Behavior of Functions
Asymptotic behavior studies how functions behave near specific points or at infinity. By analyzing dominant terms in expansions, one can simplify expressions like x sin(1/x) to find limits, focusing on leading terms that dictate the function's growth or decay.
Recommended video:
5:50Asymptotes of Hyperbolas
Related Practice
Textbook Question
Power series for derivatives
a. Differentiate the Taylor series centered at 0 for the following functions.
b. Identify the function represented by the differentiated series.
c. Give the interval of convergence of the power series for the derivative.
f(x) = eˣ
Textbook Question
Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once). Give the interval of convergence for the resulting series.
g(x) = x/(1 + x²)² using f(x) = 1/(1 + x²)
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Textbook Question
Radius and interval of convergence Determine the radius and interval of convergence of the following power series.
∑ₖ₌₀∞ (k²⁰ xᵏ)/(2k+1)!
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Textbook Question
In terms of the remainder, what does it mean for a Taylor series for a function f to converge to f?
Textbook Question
Manipulating Taylor series Use the Taylor series in Table 11.5 to find the first four nonzero terms of the Taylor series for the following functions centered at 0.
1/(1 − 2x)
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