/*! 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 11 Evaluate the following limits us... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 11

Evaluate the following limits using Taylor series. $$\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{x}$$

Short Answer

Expert verified
Question: Evaluate the limit as \(x\) approaches 0 of the expression \(\frac{e^x - e^{-x}}{x}\). Answer: The limit as \(x\) approaches 0 of the expression \(\frac{e^x - e^{-x}}{x}\) is equal to 1.

Step by step solution

01

Recall the definition of Taylor series

First, we need to recall the definition of a Taylor series. The Taylor series of a function \(f(x)\) about a point \(x=a\) is given by: $$f(x) = \sum_{n=0}^\infty \frac{f^n(a)}{n!}(x-a)^n$$ Specifically, the Taylor series expansion of the exponential function \(e^x\) about \(x=0\) is given by: $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$
02

Find Taylor series of \(e^x\) and \(e^{-x}\)

Now let's find the Taylor series expansion for both \(e^x\) and \(e^{-x}\) about \(x=0\): $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$ $$e^{-x} = \sum_{n=0}^\infty \frac{(-x)^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^nx^n}{n!}$$
03

Substitute into the expression

Now we'll substitute these Taylor series expansions of \(e^x\) and \(e^{-x}\) into the given expression, and then simplify: $$\frac{e^x - e^{-x}}{x} = \frac{\left(\sum_{n=0}^\infty \frac{x^n}{n!}\right) - \left(\sum_{n=0}^\infty \frac{(-1)^nx^n}{n!}\right)}{x}$$
04

Combine the series and cancel terms

Next, we'll combine the series and look for any cancellation of terms: $$\frac{e^x - e^{-x}}{x} = \frac{\sum_{n=0}^\infty \frac{x^n}{n!}\left(1-(-1)^n\right)}{x}$$ Notice that the series only includes odd values of n; even values are cancelled out due to the factor \(\left(1-(-1)^n\right)\) being zero for even \(n\). The series now becomes: $$\frac{e^x - e^{-x}}{x} = \frac{\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}}{x}$$
05

Simplify and take the limit

Now, we simplify this expression by replacing \(x\) in the denominator: $$\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{x} = \lim_{x \rightarrow 0} \sum_{n=0}^\infty \frac{x^{2n}}{(2n+1)!}$$ Finally, to evaluate the limit as \(x \rightarrow 0\), we just need to plug in \(x=0\) into the series: $$\lim_{x \rightarrow 0} \sum_{n=0}^\infty \frac{x^{2n}}{(2n+1)!} = \sum_{n=0}^\infty \frac{0^{2n}}{(2n+1)!}$$ For clarity, this summation only includes the first term, as the other terms in the series have a power of zero, making them vanish: $$\lim_{x \rightarrow 0} \frac{e^{x}-e^{-x}}{x} = \frac{0^0}{(2\cdot 0 + 1)!} = \frac{1}{1} = 1$$ So, the limit is equal to 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!

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

Exponential function In Section 3, we show that the power series for the exponential function centered at 0 is $$e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !}, \quad \text { for }-\infty < x < \infty$$ Use the methods of this section to find the power series for the following functions. Give the interval of convergence for the resulting series. $$f(x)=e^{-x}$$

Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers. $$\sqrt{e}$$

Replace \(x\) by \(x-1\) in the series \(\ln (1+x)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1} x^{k}}{k}\) to obtain a power series for \(\ln x\) centered at \(x=1 .\) What is the interval of convergence for the new power series?

Determine whether the following statements are true and give an explanation or counterexample. a. To evaluate \(\int_{0}^{2} \frac{d x}{1-x},\) one could expand the integrand in a Taylor series and integrate term by term. b. To approximate \(\pi / 3,\) one could substitute \(x=\sqrt{3}\) into the Taylor series for \(\tan ^{-1} x\) c. \(\sum_{k=0}^{\infty} \frac{(\ln 2)^{k}}{k !}=2\)

Exponential function In Section 3, we show that the power series for the exponential function centered at 0 is $$e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !}, \quad \text { for }-\infty < x < \infty$$ Use the methods of this section to find the power series for the following functions. Give the interval of convergence for the resulting series. $$f(x)=e^{2 x}$$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

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.