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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
11장, 문제 11.R.51

Limits by power series Use Taylor series to evaluate the following limits.


lim ₙ → ₄ ln (x - 3)/(x² - 16)

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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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.
추천 영상:
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Taylor Series

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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Geometric Series

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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