Domanda del libro di testo
Limits with a parameter Use Taylor series to evaluate the following limits. Express the result in terms of the nonzero real parameter(s).
lim ₓ→₀ (eᵃˣ − 1)/x
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Ch. 11 - Power Series
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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TOC.tableOfContents
- 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: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 11, Problema 11.4.10
Limits Evaluate the following limits using Taylor series.
lim ₓ→₀ (sin 2x)/x
Guida verificata passo dopo passo1
Recall that the Taylor series expansion of \( \sin x \) around \( x = 0 \) is given by \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \).
To find the Taylor series for \( \sin 2x \), substitute \( 2x \) into the series for \( \sin x \), giving \( \sin 2x = 2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \cdots \).
Write the expression \( \frac{\sin 2x}{x} \) using the series expansion: \( \frac{2x - \frac{(2x)^3}{3!} + \cdots}{x} \).
Simplify the fraction by dividing each term in the numerator by \( x \), resulting in \( 2 - \frac{(2x)^3}{3! x} + \cdots \).
Evaluate the limit as \( x \to 0 \) by noting that all terms containing \( x \) vanish, leaving the constant term as the limit.
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Limits
A limit describes the value that a function approaches as the input approaches a certain point. Understanding limits is fundamental in calculus for analyzing behavior near specific points, especially when direct substitution leads to indeterminate forms like 0/0.
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One-Sided Limits
Taylor Series
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 to be simplified and limits to be evaluated more easily.
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Evaluating Limits Using Taylor Series
By expanding functions into their Taylor series near the limit point, indeterminate forms can be resolved by canceling terms or simplifying expressions. This method transforms complicated limits into algebraic ones, making it easier to find the limit value.
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Pratica correlata
Domanda del libro di testo
Suppose a power series converges if |x−3|<4 and diverges if |x−3| ≥ 4. Determine the radius and interval of convergence.
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Domanda del libro di testo
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) = − 1/(1 + x)² using f(x) = 1/(1 + x)
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Domanda del libro di testo
Radius and interval of convergence Determine the radius and interval of convergence of the following power series.
∑ₖ₌₁∞ (xᵏ/kᵏ)
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Domanda del libro di testo
Representing functions by power series Identify the functions represented by the following power series.
∑ₖ₌₀∞ (xᵏ)/(2ᵏ)
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Domanda del libro di testo
{Use of Tech} Approximations with Taylor polynomials
a. Approximate the given quantities using Taylor polynomials with n = 3.
b. Compute the absolute error in the approximation, assuming the exact value is given by a calculator.
√1.06
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