Skip to main content
Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 11, Problem 11.4.8

Limits Evaluate the following limits using Taylor series.
lim ₓ→₀ (tan ⁻¹ x − x)/x³"

Verified step by step guidance
1
Recognize that the problem asks to evaluate the limit \( \lim_{x \to 0} \frac{\tan^{-1} x - x}{x^3} \) using Taylor series expansions.
Recall the Taylor series expansion of \( \tan^{-1} x \) around \( x = 0 \): \[ \tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \]
Substitute the Taylor series expansion into the numerator: \[ \tan^{-1} x - x = \left(x - \frac{x^3}{3} + \cdots \right) - x = - \frac{x^3}{3} + \cdots \]
Rewrite the original limit expression using this substitution: \[ \lim_{x \to 0} \frac{\tan^{-1} x - x}{x^3} = \lim_{x \to 0} \frac{- \frac{x^3}{3} + \cdots}{x^3} \]
Simplify the fraction by dividing each term by \( x^3 \), then evaluate the limit by letting \( x \to 0 \), which will eliminate higher order terms, leaving the coefficient of the leading term.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2m
Was this helpful?

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. For limits, Taylor expansions help identify dominant terms and simplify evaluation.
Recommended video:
08:42
Taylor Series

Inverse Tangent Function (arctan) Properties

The inverse tangent function, arctan(x), is smooth and differentiable around zero, with a known Taylor series expansion. Understanding its series helps express arctan(x) as x minus higher-order terms, which is essential for evaluating limits involving arctan(x) near zero.
Recommended video:
3:17
Inverse Tangent

Limit Evaluation Using Series Expansion

When direct substitution in a limit leads to an indeterminate form, expanding functions into their Taylor series can reveal the behavior of the numerator and denominator. By comparing the lowest-order nonzero terms, one can compute the limit accurately without complex algebraic manipulation.
Recommended video:
06:45
Intro to Series: Partial Sums
Related Practice
Textbook Question

Functions to power series Find power series representations centered at 0 for the following functions using known power series. Give the interval of convergence for the resulting series.

f(x) = 2x/(1 + x²)²

78
views
Textbook Question

Series to functions Find the function represented by the following series, and find the interval of convergence of the series. (Not all these series are power series.)


 ∑ₖ₌₀∞ e⁻ᵏˣ

40
views
Textbook Question

Series to functions Find the function represented by the following series, and find the interval of convergence of the series. (Not all these series are power series.)


∑ₖ₌₀∞(√x − 2)ᵏ

74
views
Textbook Question

Use of Tech Linear and quadratic approximation


a. Find the linear approximating polynomial for the following functions centered at the given point a.


b. Find the quadratic approximating polynomial for the following functions centered at a.


c Use the polynomials obtained in parts (a) and (b) to approximate the given quantity.


f(x)=e⁻²ˣ, a=0; approximate e⁻⁰ᐧ².

56
views
Textbook Question

Exponential function In Section 11.3, we show that the power series for the exponential function centered at 0 is


eˣ = ∑ₖ₌₀∞ (xᵏ)/k!, for −∞ < x < ∞


Use the methods of this section to find the power series centered at 0 for the following functions. Give the interval of convergence for the resulting series.


f(x) = x²eˣ

81
views
Textbook Question

Radius of convergence Find the radius of convergence for the following power series.

∑ₖ₌₁∞ (1−cos (1/2ᵏ)) xᵏ

70
views