Textbook Question
Power series from the geometric series Use the geometric series a Σₖ ₌ ₀ ∞ (x)ᵏ = 1/(1 - x), for |x| < 1, to determine the Maclaurin series and the interval of convergence for the following functions.
ƒ(x) = 1/(1 + 5x)
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.R.51
Limits by power series Use Taylor series to evaluate the following limits.
lim ₙ → ₄ ln (x - 3)/(x² - 16)
Verified step by step guidance1
First, recognize that the limit is as \(x\) approaches 4 for the expression \(\frac{\ln(x - 3)}{x^2 - 16}\). Notice that directly substituting \(x = 4\) gives \(\frac{\ln(1)}{16 - 16} = \frac{0}{0}\), an indeterminate form, so we need to use a series expansion to evaluate the limit.
Next, rewrite the denominator \(x^2 - 16\) as \((x - 4)(x + 4)\) to better understand its behavior near \(x = 4\).
Now, expand the numerator \(\ln(x - 3)\) as a Taylor series around \(x = 4\). Let \(h = x - 4\), so \(x - 3 = 1 + h\). The Taylor series for \(\ln(1 + h)\) around \(h = 0\) is \(\ln(1 + h) = h - \frac{h^2}{2} + \frac{h^3}{3} - \cdots\).
Similarly, express the denominator in terms of \(h\): \(x^2 - 16 = (4 + h)^2 - 16 = 8h + h^2\). This gives the denominator as \(8h + h^2\) near \(h = 0\).
Finally, write the original limit expression in terms of \(h\) using the expansions: \(\frac{h - \frac{h^2}{2} + \cdots}{8h + h^2}\). Simplify this expression by factoring out \(h\) from numerator and denominator, then analyze the limit as \(h \to 0\) to find the value of the limit.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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. It approximates functions near that point, allowing complex expressions to be simplified into polynomials, which are easier to analyze, especially for limits.
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Limit Evaluation Using Series
When direct substitution in a limit leads to indeterminate forms, expressing functions as power series can help. By substituting the series expansions, one can simplify the expression and find the limit by analyzing the leading terms.
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Handling Indeterminate Forms
Limits that result in forms like 0/0 require special techniques to evaluate. Using series expansions or algebraic manipulation helps resolve these indeterminate forms by revealing the behavior of numerator and denominator near the point of interest.
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Related Practice
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