/*! 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 30 Find the value of the constant \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 30

Find the value of the constant \(k\) such that. $$\lim _{x \rightarrow 2} \frac{(x+6)(x-k)}{x^{2}+x}=4$$

Short Answer

Expert verified
The value of \(k\) is \(-1\).

Step by step solution

01

Set up the Limit Equation

Given the limit problem, we want to find the value of the constant \(k\) so that: \[ \lim _{x \rightarrow 2} \frac{(x+6)(x-k)}{x^{2}+x} = 4 \] Begin by recognizing that we'll substitute \(x = 2\) to solve for \(k\).
02

Simplify the Limit Expression

Substitute \(x = 2\) into the denominator, \(x^2 + x\): \(2^2 + 2 = 6\). Since we want the limit to equal 4, solve: \[\frac{(2+6)(2-k)}{6} = 4\].
03

Solve the Equation for k

Solve the equation \(\frac{8(2-k)}{6} = 4\). Multiply both sides by 6: \(8(2-k) = 24\).
04

Isolate and Solve for k

Divide both sides by 8: \(2-k = 3\). Solve for \(k\) by subtracting 2 from both sides: \(-k = 1\). Finally, multiply by -1: \(k = -1\).

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.

Finding Constants
In calculus, finding the value of an unknown constant in a limit problem is a common task. To find the constant, follow these steps:
  • Understand the limit problem setup. You're usually asked to find a value that makes the limit of a function as it approaches a certain point equal to a given number.
  • Substitute a specific value for the variable in the equation. This helps form an equation that usually involves the unknown constant.
  • Solve the equation. Manipulate it to isolate the constant and find its value.
For example, in the given exercise, we found the constant \(k\) by setting up an equation and substituting \(x = 2\). After simplifying the expression, we solved an algebraic equation. This step-by-step approach ensures you can find the correct constant that satisfies the limit condition.
Evaluating Limits
Evaluating limits is a crucial concept in calculus that helps determine what value a function approaches as the variable approaches a certain point. Here's how you can approach evaluating limits:
  • Direct substitution: The most straightforward method, where you substitute the approaching value into the function if it doesn't cause any issues like division by zero.
  • Simplification: If direct substitution fails (e.g., leads to 0/0), simplify the expression by factoring or canceling common terms.
  • Check your work: After evaluating the limit, always ensure the calculated value matches the expected result specified in the problem.
In our exercise, substituting \(x = 2\) directly into the simplified expression allowed us to compute and confirm the desired limit equals 4. This verification assures the accuracy of our constant value for \(k\).
Rational Functions
Understanding rational functions is vital for solving calculus problems involving limits. A rational function is a fraction where both the numerator and denominator are polynomials.
  • Key characteristics include the degree of the polynomials and the values where the denominator becomes zero, known as zeros or singularities.
  • When evaluating limits, ensure the denominator does not lead to division by zero, which would require simplification or additional techniques like L'Hopital's rule.
  • Rational functions can often be simplified by factoring, which helps in canceling out terms that complicate the limit evaluation.
In the given problem, the rational function involved required careful substitution and simplification to evaluate the limit as \(x\) approached 2. Recognizing and working with these properties ensures reliable solutions to complex calculus problems.

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

Explain what is wrong with the statement. $$\ln x>\log x.$$

Are the statements true or false? Give an explanation for your answer. The function \(g(\theta)=e^{\sin \theta}\) is periodic.

The desert temperature, \(H,\) oscillates daily between \(40^{\circ} \mathrm{F}\) at 5 am and \(80^{\circ} \mathrm{F}\) at \(5 \mathrm{pm} .\) Write a possible formula for \(H\) in terms of \(t,\) measured in hours from 5 am.

Give an explanation for your answer. If \(a\) and \(b\) are positive constants, then \(y=\ln (a x+b)\) has no vertical asymptote.

For the given \(m\) and \(n\), evaluate $$\lim _{x \rightarrow 1} f(x)$$ or explain why it does not exist, where $$f(x)=\frac{(x-1)^{n}}{(x-1)^{m}}$$ \(n\) and \(m\) are positive integers with \(n>m\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Calculus

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.