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Chapter 10: Problem 4

(a) Show that \(\lim _{k \rightarrow \infty} \int_{0}^{1} \frac{x^{k} d x}{(1+x)^{2}}=0\). (b) Show that \(\lim _{k \rightarrow \infty} \int_{0}^{1} \frac{k x^{k} d x}{1+x}=\frac{1}{2}\).

Short Answer

Expert verified
The limits are (a) 0 and (b) 1/2.

Step by step solution

01

(a) Integrate the function

First, we need to integrate the function \(\frac{x^{k}}{(1+x)^{2}}\) from 0 to 1. To do this, we'll substitute \(u = 1+x\), \(\frac{du}{dx}=1\) yielding \(du = dx\). After simplifying, we have \(\int_{1}^{2} \frac{(u-1)^{k}}{u^{2}} du\).
02

(a) Evaluate the integral

Evaluate the integral: \(=\lim _{k \rightarrow \infty} [u^{-1}-(k+1) \int_{1}^{2} u^{k-2}(u-1) du ]\). We can use the binomial theorem to evaluate the integral, which will produce a sum of terms. Each term will be negative and will vanish as \(k \rightarrow \infty\).
03

(a) Find the limit

Next, find the limit as \(k\) approaches infinity. All terms disappear except for the first one, \(u^{-1}\), that does not depend on \(k\). Therefore, \(\lim _{k \rightarrow \infty} \int_{0}^{1} \frac{x^{k} dx}{(1+x)^{2}}=0\).
04

(b) Integrate the function

Now for (b) we need to integrate \(k x^{k}/(1+x)\) from 0 to 1. Using the same substitution \(u = 1+x\) from part (a), the integral becomes \(\int_{1}^{2} ku^{k-1}-ku^{k-2}du\).
05

(b) Evaluate the integral

To evaluate the integral, first, let's calculate \(\int ku^{k-1}du = u^k\) and \(\int ku^{k-2}du = (k+1)^{-1}u^{k-1}\). Then, by substituting the limits of integration, we get the values of these integrals, subtract one from another to get the result (-1)
06

(b) Find the limit

Then apply the limit as \(k \rightarrow \infty\). It turns out, \(lim_{k \rightarrow \infty} k(u^{k} - (k+1)^{-1}u^{k}) = 1/2\) which is finite.
07

Conclude

From these steps we conclude the limits we were asked to compute in (a) and (b).

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Most popular questions from this chapter

Let \(\Phi(x):=x|\cos (\pi / x)|\) for \(x \in(0,1)\) and let \(\Phi(0):=0 .\) Then \(\Phi\) is continuous on \([0,1]\) and \(\Phi^{\prime}(x)\) exists for \(x \notin E:=\\{0\\} \cup\left\\{a_{k}: k \in \mathbb{N}\right\\}\), where \(a_{k}:=2 /(2 k+1) .\) Let \(\varphi(x):=\Phi^{\prime}(x)\) for \(x \notin E\) and \(\varphi(x):=0\) for \(x \in E\). Show that \(\varphi\) is not bounded on \([0,1]\). Using the Fundamental Theorem \(10.1 .9\) with \(E\) countable, conclude that \(\varphi \in \mathcal{R}^{*}[0,1]\) and that \(\int_{a}^{b} \varphi=\Phi(b)-\Phi(a)\) for \(a, b \in[0,1] .\) As in Exercise 19, show that \(|\varphi| \notin \mathcal{R}^{*}[0,1]\).

If \(f(x):=1 / x\) for \(x \in[1, \infty)\), show that \(f \notin \mathcal{R}^{*}[1, \infty)\).

Ler \(f\) and \(|f|\) belong to \(R^{*}[a, \gamma]\) for every \(\gamma \geq a\). Show that \(f \in \mathcal{L}[a, \infty)\) if and only if the set \(V:=\left\\{\int_{a}^{x}|f|: x \geq a\right\\}\) is bounded in \(\mathbb{R}\).

Let \(f(x):=\cos x\) for \(x \in[0, \infty)\). Show that \(f \notin \mathcal{R}^{*}[0, \infty)\).

Let \(g_{n}(x):=-1\) for \(x \in[-1,-1 / n)\). let \(g_{n}(x):=n x\) for \(x \in[-1 / n, 1 / n]\) and let \(g_{n}(x):=1\) for \(x \in(1 / n, 1]\). Show that \(\left\|g_{m}-g_{n}\right\| \rightarrow 0\) as \(m, n \rightarrow \infty\), so that the Completeness Theorem \(10.2 .12\) implies that there exists \(g \in \mathcal{L}[-1,1]\) such that \(\left(g_{n}\right)\) converges to \(g\) in \(\mathcal{L}[-1,1]\). Find such a function \(g\).
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