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Chapter 3: Problem 1

Show that there cannot be more than one number \(L\) that satisfies Definition 3.1 .1

Short Answer

Expert verified
Question: Prove that if a sequence converges, its limit is unique, using Definition 3.1.1. Answer: To show that if a sequence converges, its limit is unique, we assumed the existence of two distinct limits, \(L_1\) and \(L_2\), that satisfy Definition 3.1.1. Through the triangle inequality and a derived inequality, we found that 蔚 鈮� |L_2 鈭� L_1|, which holds for all 蔚 > 0. This conclusion implies that \(L_1 = L_2\), contradicting our initial assumption. Therefore, if a sequence converges, its limit must be unique.

Step by step solution

01

Recall Definition 3.1.1 and assume the existence of two limits

Definition 3.1.1 states: A sequence \((x_n)\) converges to a limit \(L\) if for every \(\epsilon > 0\), there exists a natural number \(N\) such that for all \(n > N\), we have \(|x_n - L| < \epsilon\). Now, let's assume there are two numbers \(L_1\) and \(L_2\) that satisfy this definition for a given sequence \((x_n)\).
02

Derive the inequality for the difference between the two limits

Since both \(L_1\) and \(L_2\) satisfy Definition 3.1.1, for every \(\epsilon > 0\), there exist natural numbers \(N_1\) and \(N_2\) such that for all \(n > N_1\) we have \(|x_n - L_1| < \frac{\epsilon}{2}\) and for all \(n > N_2\) we have \(|x_n - L_2| < \frac{\epsilon}{2}\). Now, let \(N = \max\{N_1, N_2\}\). By definition, \(N \geq N_1\) and \(N \geq N_2\). Thus, for all \(n > N\), we have \(|x_n - L_1| < \frac{\epsilon}{2}\) and \(|x_n - L_2| < \frac{\epsilon}{2}\).
03

Use the triangle inequality to derive a contradiction

By the triangle inequality, we have \(|x_n - L_1| + |x_n - L_2| \geq |(x_n-L_1) - (x_n - L_2)| = |L_2 - L_1|\). Since \(|x_n - L_1| < \frac{\epsilon}{2}\) and \(|x_n - L_2| < \frac{\epsilon}{2}\) for all \(n > N\), we get: \(\frac{\epsilon}{2} + \frac{\epsilon}{2} \geq |L_2 - L_1| \Rightarrow \epsilon \geq |L_2 - L_1|\) Since the inequality above holds for all \(\epsilon > 0\), we conclude that \(L_1=L_2\), which contradicts our initial assumption that there were two distinct limits.
04

Conclude the proof

We have shown through contradiction that there cannot be more than one number \(L\) that satisfies Definition 3.1.1. Thus, if a sequence converges, its limit is unique.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniqueness of Limits
When we talk about the uniqueness of limits, we mean that a sequence can't have more than one limit. Imagine you're looking at a specific destination on a map. If a sequence of steps consistently leads to one particular spot, it can't lead to another spot at the same time.
This concept comes from Definition 3.1.1, which tells us that if a sequence converges to a limit, it does so uniquely. To illustrate this, consider a sequence \(x_n\) that converges to a limit \(L\). By definition, for every \(\epsilon > 0\), there is a point in the sequence where all further terms are so close to \(L\) that they fall within a distance \(|x_n - L| < \epsilon\).
Now, let鈥檚 say we incorrectly assume two limits, \(L_1\) and \(L_2\), exist for the same sequence. Using mathematics, it turns out that the distance between \(L_1\) and \(L_2\) has to be smaller than every possible positive number \(\epsilon\). As \(\epsilon\) can be as tiny as we like, this can only be true if \(L_1 = L_2\). This straightforward proof guarantees the uniqueness of a limit.
Convergence of Sequences
Understanding convergence is about recognizing that the terms in a sequence become consistently close to a specific number. Imagine throwing a stone into a pond and watching the ripples get closer and closer to a certain pattern 鈥� that's what convergence looks like with numbers.
For a sequence \(x_n\) to converge to a limit \(L\), the terms of the sequence must get arbitrarily close to \(L\) as you move further along the sequence. This is expressed as: for every \(\epsilon > 0\), there exists some natural number \(N\) such that whenever \(n > N\), \(|x_n - L| < \epsilon\). In other words, after a certain point in the sequence, all terms are within a tiny distance \(\epsilon\) from \(L\).
Convergence is how we know where a sequence is "heading" as the number of terms grows. It tells us about the behavior of the sequence over time, providing fundamental insights into mathematical analysis. It's like predicting the long-term destination of a moving train.
Epsilon-Delta Definition
The epsilon-delta definition is a formal mathematical way to define convergence. Don鈥檛 let the jargon intimidate you! It's actually a precise method to show just how close the terms of a sequence are to the limit.
\(\epsilon\) represents any small positive number we choose. It signifies how close we want the sequence terms to be to the limit. \(\delta\), in the context of sequences, is substituted by \(N\), a point after which all terms are close enough to the limit.
In terms of a sequence \(x_n\) converging to \(L\), the definition states that for every positive \(\epsilon\), there exists a position \(N\) such that \(n > N\) implies \(|x_n - L| < \epsilon\).
This concept is crucial when you need to verify the behavior of sequences. By fitting the terms within \(\epsilon\) distance of \(L\), we can ensure the sequence reaches its desired convergence with mathematical precision. It's like setting strict bounds to guide a path straight to the target.

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Most popular questions from this chapter

Prove: If \(f\) is locally integrable on \([a, b),\) then \(\int_{a}^{b} f(x) d x\) exists if and only if for each \(\epsilon>0\) there is a number \(r\) in \((a, b)\) such that $$ \left|\int_{x_{1}}^{x_{2}} f(t) d t\right|<\epsilon $$ whenever \(r \leq x_{1}, x_{2}

Use Theorem 3.5 .6 to show that if \(f\) is integrable on \([a, b],\) then so is \(|f|\) and, if \(f(x) \geq \rho>0(a \leq x \leq b),\) so is \(1 / f\)

Show that neither the existence nor the value of the improper integral of Definition 3.4 .3 depends on the choice of the intermediate point \(\alpha\).

Let \(u_{1}\) be positive and satisfy the differential equation $$ u^{\prime \prime}+p(x) u=0, \quad 0 \leq x<\infty $$ (a) Prove: If $$ \int_{0}^{\infty} \frac{d x}{u_{1}^{2}(x)}<\infty $$ then the function $$ u_{2}(x)=u_{1}(x) \int_{x}^{\infty} \frac{d t}{u_{1}^{2}(t)} $$ also satisfies (A), while if $$ \int_{0}^{\infty} \frac{d x}{u_{1}^{2}(x)}=\infty $$ then the function $$ u_{2}(x)=u_{1}(x) \int_{0}^{x} \frac{d t}{u_{1}^{2}(t)} $$ also satisfies \((\mathrm{A})\). (b) Prove: If (A) has a solution that is positive on \([0, \infty),\) then (A) has solutions \(y_{1}\) and \(y_{2}\) that are positive on \((0, \infty)\) and have the following properties: $$ \begin{aligned} y_{1}(x) y_{2}^{\prime}(x)-y_{1}^{\prime}(x) y_{2}(x)=1, & x>0 \\ \left[\frac{y_{1}(x)}{y_{2}(x)}\right]^{\prime} &<0, \quad x>0 \end{aligned} $$ and $$ \lim _{x \rightarrow \infty} \frac{y_{1}(x)}{y_{2}(x)}=0 $$

Suppose that \(f\) and \(g\) are integrable on \([a, b]\) and \(f(x)=g(x)\) except for \(x\) in a set of Lebesgue measure zero. Show that $$ \int_{a}^{b} f(x) d x=\int_{a}^{b} g(x) d x $$
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