/*! 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 70 Evaluate the following limits. ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 70

Evaluate the following limits. $$\lim _{\theta \rightarrow 0} \frac{\frac{1}{2+\sin \theta}-\frac{1}{2}}{\sin \theta}$$

Short Answer

Expert verified
Based on the given solution, create a short answer question: Question: Evaluate the limit of the following expression as 饾渻 approaches 0: $$\lim_{\theta\rightarrow 0}\frac{\frac{1}{2+\sin \theta}-\frac{1}{2}}{\sin \theta}$$ Answer: $$\frac{-1}{4}$$

Step by step solution

01

Rewrite the expression with a common denominator

To rewrite the expression with a common denominator, we need to consider the two fractions in the numerator: \(\frac{1}{2+\sin \theta}\) and \(\frac{1}{2}\). The common denominator for these two fractions will be \((2+\sin \theta)\times(2)\), and we will rewrite them accordingly: $$\frac{\frac{1}{2+\sin \theta}-\frac{1}{2}}{\sin \theta} = \frac{\frac{2 - (2+\sin\theta)}{2(2+\sin\theta)}}{\sin\theta} $$
02

Simplify the numerator

Now, we will simplify the numerator: $$\frac{2 - (2+\sin\theta)}{2(2+\sin\theta)} = \frac{-\sin\theta}{2(2+\sin\theta)}$$ So, the expression becomes: $$\frac{\frac{-\sin\theta}{2(2+\sin\theta)}}{\sin\theta} $$
03

Divide the expression by sin 饾渻

Next, we will divide the expression by sin 饾渻: $$\frac{-\sin\theta}{\sin\theta(2(2+\sin\theta))}$$
04

Simplify the expression further by cancelling terms

Now, we can cancel the common term \(\sin\theta\) from the numerator and denominator: $$\frac{-1}{2(2+\sin\theta)}$$
05

Evaluate the limit

Finally, we will evaluate the limit as 饾渻 approaches 0: $$\lim_{\theta\rightarrow 0}\frac{-1}{2(2+\sin\theta)} = \frac{-1}{2(2+\sin(0))} = \frac{-1}{2(2+0)} = \frac{-1}{4}$$ So, the limit $$\lim_{\theta\rightarrow 0}\frac{\frac{1}{2+\sin \theta}-\frac{1}{2}}{\sin \theta}$$ is equal to $$\frac{-1}{4}$$.

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.

Trigonometric Limits
Trigonometric limits are essential in calculus, especially when dealing with functions involving trigonometric expressions. For this exercise, we need to evaluate a limit where trigonometric functions are part of the expression, specifically with \(\sin \theta\). These limits often involve small angle approximations or identities, but here, we use division and simplification instead.

Understanding trigonometric limits often includes recognizing common limits such as:
  • \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\)
  • \(\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = 0\)
In this example, we perform algebraic manipulation to simplify the fraction and use these identities indirectly to solve the problem. Knowing these limits helps when approaching similar problems with trigonometric expressions.
Limit Evaluation
Limit evaluation is a process in calculus used to find the value that a function approaches as the input approaches a certain number. In this problem, we want to determine the behavior as \(\theta\) approaches zero. Evaluating limits often involves simplifying the expression or using strategies like L'H么pital's Rule when direct substitution isn't possible.

Key steps in this evaluation include:
  • Finding a common denominator to combine fractions and simplify the expression
  • Dividing through by terms to isolate behaviors like \(\sin \theta\)
  • Cancelling terms wisely to make the expression manageable
The final step typically involves substituting the limit value into the simplified expression. If simplification is done correctly, this will help in directly finding the limiting behavior.
Calculus Problem Solving
Calculus problem solving involves a series of logical steps that lead to understanding and solving complex problems. Each problem can often be broken down into manageable parts, just like solving the limit in this exercise.

To tackle similar problems effectively, you can:
  • Rewrite complex expressions to look simpler using algebraic identities
  • Break down the problem into smaller parts to simplify the overall expression
  • Utilize known limit properties and trigonometric identities
  • Constantly check intermediate steps to ensure accuracy in calculations
This process not only aids in solving limits but also builds a solid foundation for other areas in calculus. By following a structured approach, students can confidently solve problems and grasp the underlying concepts.

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

Let $$g(x)=\left\\{\begin{array}{ll}x^{2}+x & \text { if } x<1 \\\a & \text { if } x=1 \\\3 x+5 & \text { if } x>1.\end{array}\right.$$ a. Determine the value of \(a\) for which \(g\) is continuous from the left at 1. b. Determine the value of \(a\) for which \(g\) is continuous from the right at 1. c. Is there a value of \(a\) for which \(g\) is continuous at \(1 ?\) Explain.

Evaluate the following limits. $$\lim _{x \rightarrow \pi / 2} \frac{\sin x-1}{\sqrt{\sin x}-1}$$

Evaluate the following limits. $$\lim _{x \rightarrow 0^{+}} \frac{1-\cos ^{2} x}{\sin x}$$

A sequence is an infinite, ordered list of numbers that is often defined by a function. For example, the sequence \(\\{2,4,6,8, \ldots\\}\) is specified by the function \(f(n)=2 n\), where \(n=1,2,3, \ldots .\) The limit of such a sequence is \(\lim _{n \rightarrow \infty} f(n)\), provided the limit exists. All the limit laws for limits at infinity may be applied to limits of sequences. Find the limit of the following sequences, or state that the limit does not exist. \(\left\\{4,2, \frac{4}{3}, 1, \frac{4}{5}, \frac{2}{3}, \ldots\right\\},\) which is defined by \(f(n)=\frac{4}{n},\) for \(n=1,2,3, \ldots\)

a. Evaluate \(\lim _{x \rightarrow \infty} f(x)\) and \(\lim _{x \rightarrow-\infty} f(x),\) and then identify any horizontal asymptotes. b. Find the vertical asymptotes. For each vertical asymptote \(x=a\), evaluate \(\lim _{x \rightarrow a^{-}} f(x)\) and \(\lim _{x \rightarrow a^{+}} f(x)\). $$f(x)=\frac{\sqrt{16 x^{4}+64 x^{2}}+x^{2}}{2 x^{2}-4}$$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

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.