/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Use series to evaluate the limit... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 36

Use series to evaluate the limits. $$\lim _{x \rightarrow \infty}(x+1) \sin \frac{1}{x+1}$$

Short Answer

Expert verified
The limit evaluates to 1.

Step by step solution

01

Substitute and Simplify the Expression

The given limit is \( \lim_{x \rightarrow \infty} (x+1) \sin \frac{1}{x+1} \). Let's set \( y = x + 1 \), then as \( x \rightarrow \infty \), \( y \rightarrow \infty \) too. The limit becomes \( \lim_{y \rightarrow \infty} y \sin \frac{1}{y} \).
02

Use the Series Expansion of Sine Function

We know that the series expansion of the sine function near zero is \( \sin z = z - \frac{z^3}{6} + \mathcal{O}(z^5) \). For \( z = \frac{1}{y} \), the expression becomes \( \sin \frac{1}{y} \approx \frac{1}{y} - \frac{1}{6y^3} + \mathcal{O}\left(\frac{1}{y^5}\right) \).
03

Multiply the Series by \( y \)

Substitute the series expansion of \( \sin \frac{1}{y} \) into the expression: \( y\left( \frac{1}{y} - \frac{1}{6y^3} + \mathcal{O}\left(\frac{1}{y^5}\right) \right) = 1 - \frac{1}{6y^2} + \mathcal{O}\left(\frac{1}{y^4}\right) \).
04

Evaluate the Limit

As \( y \rightarrow \infty \), the term \( \frac{1}{6y^2} \rightarrow 0 \) and hence all higher order terms like \( \frac{1}{y^4} \rightarrow 0 \) as well. The result of the limit is simply \( 1 \). We conclude that \( \lim_{y \rightarrow \infty} (y \sin \frac{1}{y}) = 1 \), which means \( \lim_{x \rightarrow \infty} (x+1) \sin \frac{1}{x+1} = 1 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series Expansion
Series expansion is a mathematical technique that helps to represent complex functions as infinite sums of simpler terms. This is particularly useful when dealing with functions that behave nicely near certain points, providing a powerful tool to approximate and calculate difficult expressions.
  • A common series expansion is the Taylor series, which approximates functions around a particular point.
  • This involves terms of increasing powers of the variable, each multiplied by coefficients that depend on the function's derivatives.
  • It is especially useful for functions like sine, cosine, and exponentials, which have well-known expansions.
In our exercise, we use the series expansion of the sine function to break down the complicated limit expression. By expanding sine near zero, we simplify the expression significantly, allowing us to easily compute the limit at infinity.
Sine Function
The sine function is one of the fundamental trigonometric functions. It relates an angle in a right-angled triangle to the ratio of the length of the opposite side to the hypotenuse.
  • Mathematically, the sine of an angle 胃 is denoted as \( \sin \theta \).
  • The sine function is periodic with a period of \( 2\pi \), meaning it repeats its pattern every \( 2\pi \) radians.
  • Near zero, the sine function behaves very simply, which is why we can use its series expansion: \( \sin z = z - \frac{z^3}{6} + \mathcal{O}(z^5) \).
In the context of our problem, the sine function is expanded as a series around the point zero because \( \sin \frac{1}{x+1} \) simplifies to this form as \( x \rightarrow \infty \). This gives us a much simpler form to work with and helps in evaluating the limit.
Asymptotic Behavior
Asymptotic behavior describes how functions behave as the input grows indefinitely large or approaches a certain value, typically zero or infinity.
  • Understanding asymptotes helps predict the behavior of functions without needing exact values.
  • For limits involving infinity, determining the dominant terms in a function's series expansion reveals its asymptotic behavior.
  • In problems like ours, once simplified, higher-order terms approach zero as the variable goes to infinity.
In our exercise, we analyzed how \( y \sin \frac{1}{y} \) behaves as \( y \rightarrow \infty \). The asymptotic behavior showed that only the leading term of the series expansion of \( \sin \frac{1}{y} \) significantly contributes to the limit, simplifying the evaluation down to 1 as \( y \) becomes very large.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which of the series, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{n^{n}}{\left(2^{n}\right)^{2}}$$

Neither the Ratio Test nor the Root Test helps with \(p\) -series. Try them on $$ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $$ and show that both tests fail to provide information about convergence.

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=2}^{\infty} \frac{-n}{(\ln n)^{n}}$$

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=2}^{\infty} \frac{n}{(\ln n)^{(n / 2)}}$$

The series $$\sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9 !}-\frac{x^{11}}{11 !}+\cdots$$ converges to \(\sin x\) for all \(x\). a. Find the first six terms of a series for \(\cos x .\) For what values of \(x\) should the series converge? b. By replacing \(x\) by \(2 x\) in the series for \(\sin x,\) find a series that converges to \(\sin 2 x\) for all \(x\) c. Using the result in part (a) and series multiplication, calculate the first six terms of a series for \(2 \sin x \cos x .\) Compare your answer with the answer in part (b).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.