Chapter 5: Problem 2
Find a real number \(c\) and an \(N_{0} \in \mathbb{N}\) such that \(n^{3}+5
n^{2}+2 n
Short Answer
Expert verified
Choose \(c = 2\) and \(N_0 = 6\).
Step by step solution
01
Simplify the Inequality
Simplify the inequality \(n^3 + 5n^2 + 2n < cn^3\) by dividing all terms by \(n^3\): \[1 + \frac{5}{n} + \frac{2}{n^2} < c\] for \(n \geq N_0\).
02
Analyze the Limit as \(n\) Approaches Infinity
Consider the behavior of \(\frac{5}{n}\) and \(\frac{2}{n^2}\) as \(n\) becomes very large. Both \(\frac{5}{n}\) and \(\frac{2}{n^2}\) approach 0. Thus, the inequality simplifies in the limit to: \[1 < c\].
03
Choose \(c\)
Since \(1 < c\) for the inequality to hold, choose \(c\) slightly greater than 1, such as \(c = 2\). This choice satisfies \(1 < c\).
04
Determine \(N_0\)
To ensure \(1 + \frac{5}{n} + \frac{2}{n^2} < 2\), evaluate \[1 + \frac{5}{n} + \frac{2}{n^2} < 2\], which simplifies to \[\frac{5}{n} + \frac{2}{n^2} < 1\]. When \(n = 6\), \(\frac{5}{6} + \frac{2}{36} \approx 0.833 + 0.056 = 0.889\), which is indeed less than 1. Hence, choose \(N_0 = 6\).
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inequality Simplification
Simplifying inequalities is like breaking a complex problem into smaller, more manageable pieces, allowing us to better handle the situation. When faced with the inequality \(n^3 + 5n^2 + 2n < cn^3\), the goal is to simplify the terms in a way that makes it easier to solve. To do this, we divide every term in the inequality by the highest power of \(n\) present, which is \(n^3\). This process is beneficial because it reduces each term to something simpler and often reveals the core idea of the inequality in a cleaner mathematical expression. As a result, our inequality becomes \[1 + \frac{5}{n} + \frac{2}{n^2} < c\]. This step eliminates the variable's highest power and suggests that for very large \(n\), values of \(\frac{5}{n}\) and \(\frac{2}{n^2}\) diminish in significance, making it easier to assess the inequality.
Limit Behavior
The behavior of a function as it approaches a limit helps us understand long-term trends, such as what happens as values become very large or very small. Here, our interest is in what happens as \(n\), a natural number, grows infinitely large. By examining \(\frac{5}{n}\) and \(\frac{2}{n^2}\) as \(n\) gets larger, we find that these terms become very small. Specifically, - \(\frac{5}{n}\) approaches zero because the numerator stays constant while the denominator grows.- \(\frac{2}{n^2}\) also approaches zero, but at a faster rate since \(n^2\) grows even more rapidly than \(n\) alone.With both terms approaching zero as \(n\) increases, the inequality simplifies to \(1 < c\). This limit behavior is crucial as it shows that the inequality simplifies in a straightforward manner when concentrating on the largest term, \(1\), and ignoring the diminishing fractions.
Natural Numbers
Natural numbers, denoted \(\mathbb{N}\), are a fundamental concept in mathematics. They include all positive integers starting from 1 (\{1, 2, 3, 4, \ldots\}). They are often used in mathematical problems dealing with whole, countable units. In the exercise, the problem defines \(n\) as a natural number, which means it must be a positive integer. This influences the concepts of limits and inequalities since we only consider discrete values increasing without bound.When selecting \(N_0\) in step 4, we ensure the inequality holds for all \(n \geq N_0\). In this case, selecting \(N_0 = 6\) ensures that any \(n\) greater than or equal to 6 will satisfy the inequality due to the decreasing influence of terms \(\frac{5}{n}\) and \(\frac{2}{n^2}\). We test this with \(n = 6\) as an anchor point, confirming the conditions are met with these fixed natural numbers.