/*! 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 13 Calculate. $$\lim _{x \rightar... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 13

Calculate. $$\lim _{x \rightarrow 0} \frac{x+\sin \pi x}{x-\sin \pi x}$$

Short Answer

Expert verified
The short answer is: \(\lim _{x \rightarrow 0} \frac{x+\sin \pi x}{x-\sin \pi x} = \frac{1 + \pi}{1 - \pi}\).

Step by step solution

01

Identify the indeterminate form

Before applying L'H么pital's rule, we should first find the limit of the numerator and denominator as x approaches 0. \[\lim_{x \rightarrow 0}(x + \sin(\pi x)) = 0\] \[\lim_{x \rightarrow 0}(x - \sin(\pi x)) = 0\] The limit of both expressions is 0. Given that, we have the indeterminate form \(\frac{0}{0}\).
02

Apply L'H么pital's rule

L'H么pital's rule tells us that if we have a limit of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can find the limit by taking the derivative of the numerator and the derivative of the denominator, and then finding the limit of the new fraction. First, let's find the derivatives of the numerator and denominator: \[\frac{d}{dx}(x+\sin(\pi x)) = 1+\pi \cos(\pi x)\] \[\frac{d}{dx}(x-\sin(\pi x)) = 1-\pi \cos(\pi x)\] Now, we can plug these into our limit expression, and find the limit of the new fraction: \[\lim_{x \rightarrow 0} \frac{1 + \pi \cos(\pi x)}{1 - \pi \cos (\pi x)}\]
03

Evaluate the limit of the new fraction

Lastly, we need to find the limit of the new fraction as x approaches 0: \[\lim_{x \rightarrow 0} \frac{1 + \pi \cos(\pi x)}{1 - \pi \cos (\pi x)} = \frac{1 + \pi \cos(\pi \cdot 0)}{1 - \pi \cos (\pi \cdot 0)} = \frac{1 + \pi}{1 - \pi}\] As there is no indeterminate form present, we can conclude that the original limit of the function evaluates to: \[\lim _{x \rightarrow 0} \frac{x+\sin \pi x}{x-\sin \pi x} = \frac{1 + \pi}{1 - \pi}\]

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.

Indeterminate Forms
Indeterminate forms are expressions in calculus that do not immediately lead to a clear limit or value. These forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \infty - \infty\, among others. When you encounter such expressions, it means you cannot directly compute the limit, and you'll need additional techniques to find the solution.
  • A common scenario is when both the numerator and the denominator of a fraction approach zero, forming the indeterminate form \(\frac{0}{0}\).
  • L'H么pital's Rule is a common method used to handle \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) scenarios by differentiating the numerator and denominator separately and then finding the limit of the resulting fraction.
  • Indeterminate forms signal a need for deeper analysis, whereas determinate forms straightforwardly lead to the final value without these intermediate steps.
Calculus Limits
The concept of limits is fundamental in calculus, allowing us to understand the behavior of functions as they approach a specific point.
When calculating limits, especially those involving indeterminate forms, it is crucial to analyze the behavior of both the numerator and the denominator separately.
  • Limits describe the value that a function approaches as the input approaches some value. It lays the groundwork for defining derivatives and integrals.
  • In scenarios with indeterminate forms or where functions involve complex expressions, employing limit properties and rules like L'H么pital's Rule can simplify the problem.
  • Confirming both parts of a fraction approach zero is key in identifying the need to use L'H么pital's Rule.

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 comparison test (11.7.2) to determine whether the integral converges. $$\int_{1}^{\infty} \frac{x}{\sqrt{1+x^{5}}} d x$$

Use comparison test (11.7.2) to determine whether the integral converges. $$\int_{0}^{\infty}\left(1+x^{5}\right)^{-1 / 6} d x$$

Use mathematical induction to prove the following assertions. If \(a_{1}=3\) and \(a_{n+1}=a_{n}+5,\) then \(a_{n}=5 n-2\).

Let \(S\) be a bounded set of real numbers and suppose that lub \(S=\mathrm{glb} S .\) What can you conclude about \(S ?\)

Use comparison test (11.7.2) to determine whether the integral converges. $$\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

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.