Problem 26 - A Course in Calculus and Real Analysis

Question

Find \(\lim _{n \rightarrow \infty} \int_{0}^{1} \frac{n x^{n-1}}{1+x} d x .\) (Hint: Proposition 6.28.)

Step-by-Step Solution

Answer

The limit of the given integral as n approaches infinity is 0. We used Proposition 6.28, analyzed the pointwise limit of the function, and proved the uniform convergence of the sequence of functions to arrive at this conclusion.

Step 1: Express the given function as a sequence of continuous functions

Define the function \(f_n(x) = \frac{n x^{n-1}}{1 + x}\). We want to prove that this sequence of functions converges uniformly to some continuous function f(x) on the interval [0,1]. If we can prove this, then we can apply Proposition 6.28.

Step 2: Determine the pointwise limit of the function

We need to find the pointwise limit of the function \(f_n(x)\). The pointwise limit is given by the limit of the function as n approaches infinity for a given x value: \[ f(x) = \lim_{n\rightarrow \infty} f_n(x) = \lim_{n\rightarrow \infty} \frac{n x^{n-1}}{1+x}. \] Let's analyze the limit in two cases: Case 1: x = 0, then the limit will become \[ f(0) = \lim_{n\rightarrow \infty} \frac{n \cdot 0^{n-1}}{1+0} = 0, \] since the numerator will be 0. Case 2: \(0 < x \leq 1\), then we can use the following inequality, which holds for any n: \[ nx^{n-1} \leq nx^n \leq n, \] since \(0 < x \leq 1\), any x raised to a power larger than 1 is always smaller than or equal to x. So, we can use the squeeze theorem for the limit: \[ 0 \leq \lim_{n\rightarrow \infty} \frac{n x^{n-1}}{1 + x} \leq \lim_{n\rightarrow \infty} \frac{n}{1 + x}, \] but the last limit is clearly zero, thus \[ f(x) = \lim_{n\rightarrow \infty} \frac{n x^{n-1}}{1 + x} = 0. \]

Step 3: Prove uniform convergence of the sequence of functions

We have found the pointwise limit for the function to be f(x) = 0. Since f(x) is continuous on the interval [0,1], we only need to show that the sequence \(f_n(x) \) converges uniformly to f(x) on this interval. We'll use the following definition of uniform convergence: For every \(\varepsilon > 0\), there exists some N such that for all n \(\geq\) N, and for all x in [0,1], \[ |f_n(x) - f(x)| < \varepsilon. \] Since we have already found that \(f(x) = 0\), this inequality becomes: \[ |f_n(x)| = \left|\frac{n x^{n-1}}{1+x}\right| < \varepsilon. \] From the analysis in Step 2, we already know that the limit of this function is 0 for all x in the interval, which means that for a sufficiently large value of n, the inequality will hold true for all x in [0,1]. Therefore, the sequence of functions \(f_n(x)\) converges uniformly to f(x) on the interval [0,1].

Step 4: Apply Proposition 6.28 to find the limit of the integral

Now, we can apply Proposition 6.28 to our problem: \[ \lim_{n \rightarrow \infty} \int_{0}^{1} \frac{n x^{n-1}}{1+x} dx = \int_{0}^{1} f(x) dx. \] As we found in Step 2 that f(x) = 0, the integral becomes: \[ \int_{0}^{1} 0 dx = 0. \] Therefore, the limit of the given integral as n approaches infinity is 0.

Key Concepts

Pointwise Convergence

Pointwise convergence is a fundamental concept when analyzing sequences of functions. It involves examining the behavior of a sequence of functions at each point within a specific domain. For the given exercise, we define a sequence of functions \( f_n(x) = \frac{n x^{n-1}}{1+x} \). To determine if this sequence converges pointwise, we look at the limit of \( f_n(x) \) as \( n \to \infty \) for each fixed \( x \):

• At \( x = 0 \), \( f_n(0) = \frac{n \cdot 0^{n-1}}{1+0} = 0 \). Here, the sequence converges to 0.
• For \( 0 < x \leq 1 \), applying inequality and taking the limit, we also find \( f(x) = 0 \).

This shows that the sequence converges to the function \( f(x) = 0 \) pointwise over the interval \([0,1]\). This intuition is crucial because achieving pointwise convergence is the first step before exploring other types of convergence, such as uniform convergence.

Squeeze Theorem

The squeeze theorem is a powerful tool in calculus, used to determine the limit of a function by 'squeezing' it between two other functions with known limits. It is particularly handy when direct evaluation or algebraic manipulation of the given function is complex or unwieldy. In step 2 of our solution, the squeeze theorem was used to find the limit of \( f_n(x) \) for \( 0 < x \leq 1 \). Let's illustrate this with the inequality applied:

• We have: \( 0 \leq nx^{n-1} \leq n \)
• And dividing by \( 1+x \): \( 0 \leq \frac{n x^{n-1}}{1 + x} \leq \frac{n}{1 + x} \)

Since \( \lim_{n \to \infty} \frac{n}{1+x} = 0 \), the squeeze theorem tells us that \( \lim_{n \to \infty} \frac{n x^{n-1}}{1 + x} = 0 \) as well. This method makes finding limits much simpler, provided we can establish a valid inequality.

Proposition 6.28

Proposition 6.28 deals with uniform convergence and the interchange of limits and integrals. It states that if a sequence of functions converges uniformly to a function \( f(x) \) on an interval, then the integral of the limit function is the limit of the integrals: \[ \lim_{n \rightarrow \infty} \int_{a}^{b} f_n(x) \, dx = \int_{a}^{b} \lim_{n \rightarrow \infty} f_n(x) \, dx \]In our exercise, after establishing that \( f_n(x) = \frac{n x^{n-1}}{1+x} \) converges uniformly to \( f(x) = 0 \) on \([0,1]\), proposition 6.28 allows us to take the limit of the integral:

• \( \lim_{n \rightarrow \infty} \int_{0}^{1} \frac{n x^{n-1}}{1+x} dx = \int_{0}^{1} 0 \, dx = 0 \)

Understanding proposition 6.28 is pivotal in problems dealing with the convergence of integrals, enabling us to effortlessly equate the integral of a limit function with the limit of its integrals.