/*! 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 67 Use Taylor series to evaluate th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 67

Use Taylor series to evaluate the following limits. Express the result in terms of the parameter(s). $$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}$$

Short Answer

Expert verified
Answer: The value of the limit \(\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}\) is \(\frac{a}{b}\).

Step by step solution

01

Taylor series expansion for \(\sin(ax)\)

To find the Taylor series expansion for \(\sin(ax)\), we need to recall the Taylor series for \(\sin(x)\) and substitute \(x\) with \(ax\), which is: $$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$ So for \(\sin(ax)\), we have: $$\sin(ax) = ax - \frac{(ax)^3}{3!} + \frac{(ax)^5}{5!} - \frac{(ax)^7}{7!} + \cdots$$
02

Taylor series expansion for \(\sin(bx)\)

Similarly, we find the Taylor series expansion for \(\sin(bx)\) by substituting \(x\) with \(bx\) in the Taylor series for \(\sin(x)\): $$\sin(bx) = bx - \frac{(bx)^3}{3!} + \frac{(bx)^5}{5!} - \frac{(bx)^7}{7!} + \cdots$$
03

Divide the two Taylor series

Now we divide the Taylor series of \(\sin(ax)\) by the Taylor series of \(\sin(bx)\). With this division, we get the new series: $$\frac{\sin(ax)}{\sin(bx)} = \frac{ax - \frac{(ax)^3}{3!} + \frac{(ax)^5}{5!} - \frac{(ax)^7}{7!} + \cdots}{bx - \frac{(bx)^3}{3!} + \frac{(bx)^5}{5!} - \frac{(bx)^7}{7!} + \cdots}$$
04

Evaluate the limit

The next step is to evaluate the limit as \(x\) approaches \(0\): $$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x} = \lim _{x \rightarrow 0}\frac{ax - \frac{(ax)^3}{3!} + \frac{(ax)^5}{5!} - \frac{(ax)^7}{7!} + \cdots}{bx - \frac{(bx)^3}{3!} + \frac{(bx)^5}{5!} - \frac{(bx)^7}{7!} + \cdots}$$
05

Simplify and express the result in terms of the parameters

Observe that each term in the series for \(\frac{\sin(ax)}{\sin(bx)}\) contains a power of \(x\). This means that as \(x\) approaches \(0\), the higher order terms (those with larger powers of \(x\)) will become negligible compared to the lower order terms. Thus, we can approximate the limit by considering only the first terms in the two series (i.e., those with the lowest powers of \(x\)). This gives us: $$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x} \approx \lim _{x \rightarrow 0}\frac{ax }{bx } = \frac{a}{b}$$ So the result of the limit is: $$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x} = \frac{a}{b}$$

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!

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

Use Taylor series to evaluate the following limits. Express the result in terms of the parameter(s). $$\lim _{x \rightarrow 0} \frac{e^{a x}-1}{x}$$

What is the minimum order of the Taylor polynomial required to approximate the following quantities with an absolute error no greater than \(10^{-3}\) ? (The answer depends on your choice of a center.) $$e^{-0.5}$$

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. In most cases you do not need to use the definition of the Taylor series coefficients. b. If possible, determine the radius of convergence of the series. $$f(x)=\tan x$$

Use properties of power series, substitution, and factoring of constants to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. Use the Taylor series. $$(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+\cdots, \text { for }-1 < x < 1$$ $$(1+4 x)^{-2}$$

Consider the following common approximations when \(x\) is near zero. a. Estimate \(f(0.1)\) and give the maximum error in the approximation. b. Estimate \(f(0.2)\) and give the maximum error in the approximation. $$f(x)=\tan ^{-1} x \approx x$$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

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.