/*! 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 28 Find \(\lim _{n \rightarrow \inf... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 28

Find \(\lim _{n \rightarrow \infty} \int_{0}^{1} \frac{n x^{n-1}}{1+x} d x .\) (Hint: Proposition 6.25.)

Short Answer

Expert verified
Using Proposition 6.25, we compared the given integral with the simpler integral \(g_n(x) = n x^{n-1}\) and found the limit of the given integral as n approaches infinity: \(\lim _{n \rightarrow \infty} \int_{0}^{1} \frac{n x^{n-1}}{1+x} d x = \boxed{1}\).

Step by step solution

01

Identify the functions \(f_n(x)\) and \(g_n(x)\) for Proposition 6.25

In this case, the function we are given is \(\frac{n x^{n-1}}{1+x}\). We will try to find another function \(g_n(x)\) such that \(0 \leq f_{n}(x) \leq g_{n}(x)\) for all \(n\) and \(x\). A suitable function to compare with is \(g_n(x) = n x^{n-1}\), since we know that \(\frac{1}{1+x} \leq 1\) for all \(x \in [0, 1]\). Now we need to compute the integrals of \(f_n(x)\) and \(g_n(x)\) and find the limit using Proposition 6.25.
02

Compute the integrals of \(f_n(x)\) and \(g_n(x)\)

First, let's compute the integral of \(f_n(x)\): \[\int_{0}^{1} \frac{n x^{n-1}}{1+x} dx\] Next, we compute the integral of \(g_n(x)\): \[\int_{0}^{1} n x^{n-1} dx\] To find the integral of \(g_n(x)\), we can use the power rule for integration, which gives us: \[\int n x^{n-1} dx = x^n\] Now we need to evaluate this integral from 0 to 1: \[[x^n]_0^1 = 1^n - 0^n = 1 - 0 = 1\] So, the limit of the integral of \(g_n(x)\) as n approaches infinity is: \[\lim_{n \rightarrow \infty} \int_{0}^{1} n x^{n-1} dx = 1\]
03

Apply Proposition 6.25

Now that we have found the limits of the integrals of \(f_n(x)\) and \(g_n(x)\), we can apply Proposition 6.25: \[\lim_{n \rightarrow \infty} \int_{0}^{1} \frac{f_{n}(x)}{g_{n}(x)} dx = 1\] Therefore, the limit of the given integral as n approaches infinity is: \[\lim _{n \rightarrow \infty} \int_{0}^{1} \frac{n x^{n-1}}{1+x} d x = \boxed{1}\]

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.

Real Analysis
Real analysis is a foundational branch of mathematics dealing with real numbers and real-valued functions. It rigorously defines concepts such as sequences, series, limits, continuity, and differentiation, which are essential in understanding calculus and higher-level mathematics. In the context of our problem, real analysis provides the theoretical framework for manipulating the limit of an integral as the index of the sequence goes to infinity, which is a common scenario in evaluating the behavior of functions as they approach some bound or infinity.
Sequence of Functions

In real analysis, a sequence of functions is a series of functions \( f_n \) that are indexed by a natural number \( n \). Typically, we are interested in the behavior of such sequences as \( n \) grows larger - either their convergence to a specific function or their divergence. In the exercise, \( f_n(x) = \frac{n x^{n-1}}{1+x} \) forms a sequence of functions indexed by \( n \). As part of the solution, we investigate the limit of the integral of these functions as \( n \) approaches infinity.

Power Rule for Integration

The power rule is a fundamental technique used in calculus to compute the integral of a power function. For any real number \( n \) not equal to -1, the power rule states that you can integrate \( x^n \) by adding 1 to the exponent and then dividing by the new exponent, thus: \[ \int x^n dx = \frac{1}{n+1} x^{n+1} \] In our solution, applying the power rule allows us to easily integrate \( g_n(x) = n x^{n-1} \) to find out that its integral from 0 to 1 is equal to 1, for every positive integer \( n \).

Dominated Convergence Theorem

The Dominated Convergence Theorem is an important result in real analysis, particularly useful when dealing with sequences of functions and their integrals. It states that if we have a sequence of functions that converge almost everywhere to a function \( f \) and are dominated by an integrable function \( g \) meaning for all \( n \) the absolute value of \( f_n \) is less than or equal to \( g \)鈥攖hen the integral of the limit function \( f \) is equal to the limit of the integrals of \( f_n \)s. This concept is implicit in our problem, as we ultimately want to take the limit inside the integral sign. By considering \( g_n(x) \) as being a dominant function for the sequence \( f_n(x) \) in this case, we apply the theorem to deduce that the sequence of integrals converges to the same value as the integral of the limit function.

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

If \(x:=\int_{0}^{y} \frac{d t}{\sqrt{1+t^{2}}}\), find \(\frac{d^{2} y}{d x^{2}}\)

Let \(f:[a, b] \rightarrow \mathbb{R}\) be integrable and for \(n \in \mathbb{N}\), let \(P_{n}\) be a partition of \([a, b]\) such that \(U\left(P_{n}, f\right)-L\left(P_{n}, f\right) \rightarrow 0 .\) Show that \(U\left(P_{n}, f\right) \rightarrow \int_{a}^{b} f(x) d x\) \(L\left(P_{n}, f\right) \rightarrow \int_{a}^{b} f(x) d x\), and also \(S\left(P_{n}, f\right) \rightarrow \int_{a}^{b} f(x) d x\), where \(S\left(P_{n}, f\right)\) is a Riemann sum for \(f\) corresponding to \(P_{n} .\) (Compare Proposition \(6.5\) and Lemma 6.30.)

Let \(f, g:[a, b] \rightarrow \mathbb{R}\) be bounded functions. Show that $$ L(f)+L(g) \leq L(f+g) \quad \text { and } \quad U(f+g) \leq U(f)+U(g) $$ Hence conclude that if \(f\) and \(g\) are integrable, then so is \(f+g\), and the Riemann integral of \(f+g\) is equal to the sum of the Riemann integrals of \(f\) and \(g\).

Let \(c \in[a, b]\) and \(f:[a, b] \rightarrow \mathbb{R}\) be given by $$ f(x):=\left\\{\begin{array}{ll} 0 & \text { if } a \leq x \leq c \\ 1 & \text { if } c

Let \(f:[a, \infty) \rightarrow \mathbb{R}\) be a bounded function such that \(f\) is integrable on \([a, x]\) for every \(x \geq a\). Let \(F(x):=\int_{a}^{x} f(t) d t\) for \(x \geq a\). Show that \(F\) is uniformly continuous on \([a, \infty)\).
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Decision Maths

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.