Skip to main content
Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 11 - Power SeriesProblema 11.4.16
Capitolo 11, Problema 11.4.16

Limits Evaluate the following limits using Taylor series.
lim ₓ→₄ (x² 16)/(ln (x 3)}

Guida verificata passo dopo passo
1
First, rewrite the limit expression clearly: \(\lim_{x \to 4} \frac{x^2 - 16}{\ln(x - 3)}\).
Recognize that as \(x\) approaches 4, the numerator \(x^2 - 16\) approaches \(4^2 - 16 = 0\), and the denominator \(\ln(x - 3)\) approaches \(\ln(1) = 0\), so this is an indeterminate form \(\frac{0}{0}\) suitable for applying Taylor series expansions.
Expand the numerator \(x^2 - 16\) around \(x = 4\) using the Taylor series (or simply use the linear approximation): \(x^2 - 16 = (4)^2 - 16 + 2 \cdot 4 (x - 4) + \cdots = 0 + 8(x - 4) + \cdots\).
Expand the denominator \(\ln(x - 3)\) around \(x = 4\). Since \(x - 3\) approaches 1, use the expansion of \(\ln(1 + h)\) where \(h = x - 4\): \(\ln(x - 3) = \ln(1 + (x - 4)) = (x - 4) - \frac{(x - 4)^2}{2} + \cdots\).
Substitute these expansions back into the limit expression and simplify by canceling common factors, then evaluate the limit by taking \(x \to 4\) (or equivalently \(h \to 0\)).

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
4m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Limits and Limit Evaluation

Limits describe the behavior of a function as the input approaches a particular value. Evaluating limits helps determine the function's value near points where direct substitution may be undefined or indeterminate, such as 0/0 or ∞/∞ forms.
Video consigliato:
05:50
One-Sided Limits

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 simplification of complex expressions to evaluate limits or analyze behavior.
Video consigliato:
08:42
Taylor Series

Handling Indeterminate Forms Using Series

When direct substitution in limits results in indeterminate forms like 0/0, expanding numerator and denominator into Taylor series helps identify leading terms. This approach simplifies the limit evaluation by canceling common factors and revealing the limit's value.
Video consigliato:
Percorso guidato
06:45
Intro to Series: Partial Sums
Pratica correlata
Domanda del libro di testo

L'Hôpital's Rule by Taylor series Suppose f and g have Taylor series about the point a.

a. If f(a) = g(a) = 0 and g′(a) ≠ 0, evaluate lim ₓ→ₐ f(x)/g(x) by expanding f and g in their Taylor series. Show that the result is consistent withl’Hôpital’s Rule.

b. If f(a) = g(a) =f′(a) = g′(a) = 0 and g′′(a) ≠ 0, evaluate lim ₓ→ₐ f(x)/g(x) by expanding f and g in their Taylor series. Show that the result is consistent with two applications of 1'Hôpital's Rule.

55
views
Domanda del libro di testo

Limits Evaluate the following limits using Taylor series.

lim ₓ→∞ x(e¹/ˣ − 1)

85
views
Domanda del libro di testo

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) = sin x, a = π/2

59
views
Domanda del libro di testo

Radius and interval of convergence Determine the radius and interval of convergence of the following power series.

∑ₖ₌₀∞ k(x−1)ᵏ

71
views
Domanda del libro di testo

Approximating real numbers Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers.

√e

68
views
Domanda del libro di testo

Does the accuracy of an approximation given by a Taylor polynomial generally increase or decrease with the order of the approximation? Explain.

84
views