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Chapter 6: Problem 14

A joint limit (see above) does not mean the iterated limits even exist. Consider \(x_{n, m}:=\) \(\frac{(-1)^{n+m}}{\min [n, m]}\) a) Show that for no \(n\) does \(\lim _{m \rightarrow \infty} x_{n, m}\) exist, and for no \(m\) does \(\lim _{n \rightarrow \infty} x_{n, m}\) exist. So neither \(\lim _{n \rightarrow \infty} \lim _{n \rightarrow \infty} x_{n, m}\) nor \(\lim _{m \rightarrow \infty} \lim _{n \rightarrow \infty} x_{n, m}\) makes any sense at all. b) Show that the joint limit of \(\left\\{x_{n-m}\right\\}\) exists and is \(0 .\)

Short Answer

Expert verified
Iterated limits do not exist for any fixed \(n\) or \(m\), but the joint limit exists and is 0.

Step by step solution

01

Analyze the Sequence for Part (a)

The sequence given is \(x_{n,m} = \frac{(-1)^{n+m}}{\min[n,m]}\). We need to analyze the behavior of this sequence as \(m \to \infty\) for fixed \(n\), and as \(n \to \infty\) for fixed \(m\).
02

Evaluate Limit as m goes to Infinity

For a fixed \(n\), consider \(\lim_{m \to \infty} x_{n,m} = \lim_{m \to \infty} \frac{(-1)^{n+m}}{\min[n,m]}\). Since \(\min[n,m] = n\) for \(m \geq n\), \(x_{n,m} = \frac{(-1)^{n+m}}{n}\). Because \((-1)^{n+m}\) oscillates between \(-1\) and \(1\), the limit does not exist.
03

Evaluate Limit as n goes to Infinity

For a fixed \(m\), consider \(\lim_{n \to \infty} x_{n,m} = \lim_{n \to \infty} \frac{(-1)^{n+m}}{\min[n,m]}\). Here, \(\min[n,m] = m\) for \(n \geq m\), leading to \(x_{n,m} = \frac{(-1)^{n+m}}{m}\). Again, since \((-1)^{n+m}\) oscillates between \(-1\) and \(1\), the limit does not exist.
04

Conclude Non-existence of Iterated Limits

Since neither \(\lim_{m \to \infty}\) nor \(\lim_{n \to \infty}\) yields a defined limit, the iterated limits \(\lim_{n \to \infty} (\lim_{m \to \infty} x_{n,m})\) and \(\lim_{m \to \infty} (\lim_{n \to \infty} x_{n,m})\) do not make sense.
05

Analyze Joint Limit for Part (b)

To consider the joint limit, evaluate \(x_{n,m}\) as both \(n\) and \(m\) approach infinity. Rewriting \(x_{n,m} = \frac{(-1)^{n+m}}{\min[n,m]}\), for \(n = m\), \((-1)^{n+m} = 1\) and \(\min[n,m] = n = m\). Thus \(x_{n,n} = \frac{1}{n}\), which tends to 0 as \(n \to \infty\).
06

Conclusion of Joint Limit Existence

For any sequence where \(n\) and \(m\) approach infinity simultaneously, \((-1)^{n+m}\) may oscillate, but \(|\frac{(-1)^{n+m}}{\min[n,m]}|\) remains \(\leq \frac{1}{\min[n,m]}\). This tends to zero, implying the joint limit exists and is 0.

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Key Concepts

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

iterated limits
The concept of iterated limits involves calculating the limit of a function or a sequence by fixing one variable and letting it tend towards infinity, after which the second variable is taken to infinity. In the context of the given exercise, this process involves taking the limit of the sequence \( x_{n,m} = \frac{(-1)^{n+m}}{\min[n,m]} \) first by fixing \( n \) and letting \( m \rightarrow \infty \) or vice versa.Though it might seem straightforward, iterated limits don't always yield meaningful results. As observed in the exercise, due to the oscillatory nature of \((-1)^{n+m}\), the values alternate between
  • -1
  • 1
The denominator being finite prevents this oscillation from settling at any particular value. Therefore, such limits do not exist in this scenario, and the iterated limits \( \lim_{n \to \infty} \lim_{m \to \infty} x_{n,m} \) or \( \lim_{m \to \infty} \lim_{n \to \infty} x_{n,m} \) are undefined.
sequence analysis
Sequence analysis is a critical tool in understanding the behavior of sequences as they progress towards infinity. In the given sequence \( x_{n,m} = \frac{(-1)^{n+m}}{\min[n,m]} \), understanding the pattern is essential.When breaking down this sequence,
  • Analyzing by first fixing \( n \) shows that as \( m \to \infty \), the sequence remains undefined due to oscillations.
  • Similarly, fixing \( m \) lets us know that as \( n \to \infty \), the sequence oscillates, also remaining undefined in terms of a limit.
Through sequence analysis, it becomes evident that neither pathway for these variables results in a convergent limit. Moving forward with this understanding helps identify when a sequence might converge inherently and when it might not without further analysis.
real analysis
In real analysis, we commonly investigate sequences and their convergence to determine meaningful limits. Real analysis helps to understand behavior such as the joint limit present in this exercise. By examining both variables \( n \) and \( m \) approaching infinity simultaneously, the sequence \( x_{n,m} = \frac{(-1)^{n+m}}{\min[n,m]} \) simplifies. For instance, for \( n = m \), \( (-1)^{n+m} \) becomes \( 1 \) because \( n+m \) is even, and the sequence becomes \( \frac{1}{n} \), which clearly tends towards 0.This approach highlights the notion of a joint limit, which can exist even if iterated limits may not. By closely studying the conditions of convergence within real number bounds, real analysis allows for a precise and insightful evaluation of sequences to determine the convergence behavior, as demonstrated by the solution reaching the joint limit of 0 in this exercise.

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Most popular questions from this chapter

Suppose \(f_{n}: S \rightarrow \mathbb{R}\) are functions that converge uniformly to \(f: S \rightarrow \mathbb{R} .\) Suppose \(A \subset S .\) Show that the sequence of restrictions \(\left\\{\left.f_{n}\right|_{A}\right\\}\) converges uniformly to \(\left.f\right|_{A}\).

Let \(\left\\{f_{n}\right\\},\left\\{g_{n}\right\\}\) and \(\left\\{h_{n}\right\\}\) be sequences of functions on \([a, b] .\) Suppose \(\left\\{f_{n}\right\\}\) and \(\left\\{h_{n}\right\\}\) converge uniformly to some function \(f:[a, b] \rightarrow \mathbb{R}\) and suppose \(f_{n}(x) \leq g_{n}(x) \leq h_{n}(x)\) for all \(x \in[a, b] .\) Show that \(\left\\{g_{n}\right\\}\) converges uniformly to \(f\).

A common type of equation one encounters are linear first order differential equations, that is equations of the form $$ y^{\prime}+p(x) y=q(x), \quad y\left(x_{0}\right)=y_{0} . $$ Prove Picard's theorem for linear equations. Suppose I is an interval, \(x_{0} \in I,\) and \(p: I \rightarrow \mathbb{R}\) and \(q: I \rightarrow \mathbb{R}\) are continuous. Show that there exists a unique differentiable \(f: I \rightarrow \mathbb{R},\) such that \(y=f(x)\) satisfies the equation and the initial condition. Hint: Assume existence of the exponential function and use the integrating factor formula for existence of \(f\) (prove that it works): $$ f(x):=e^{-\int_{x_{0}}^{x} p(s) d s}\left(\int_{x_{0}}^{x} e^{\int_{x_{0}}^{t} p(s) d s} q(t) d t+y_{0}\right) $$

Find a sequence of Riemann integrable functions \(f_{n}:[0,1] \rightarrow \mathbb{R}\) such that \(\left\\{f_{n}\right\\}\) comverges to zero pointwise, and such that a) \(\left\\{\int_{0}^{1} f_{n}\right\\}_{n=1}^{\infty}\) increases without bound. b) \(\left\\{\int_{0}^{1} f_{n}\right\\}_{n=1}^{\infty}\) is the sequence \(-1,1,-1,1,-1,1, \ldots\) It is possible to define a joint limit of a double sequence \(\left\\{x_{n, m}\right\\}\) of real numbers (that is a function from \(\mathbb{N} \times \mathbb{N}\) to \(\mathbb{R}\) ). We say \(L\) is the joint limit of \(\left\\{x_{n, m}\right\\}\) and write $$ \lim _{n \rightarrow \infty} x_{n, m}=L, \quad \text { or } \quad \lim _{(n, m) \rightarrow \infty} x_{n, m}=L $$ if for every \(\varepsilon>0,\) there exists an \(M\) such that if \(n \geq M\) and \(m \geq M,\) then \(\left|x_{n, m}-L\right|<\varepsilon\).

Suppose F: \(I \times J \rightarrow \mathbb{R}\) is a function that is continuous in the first variable, that is, for any fixed y the function that takes \(x\) to \(F(x, y)\) is continuous. Further, suppose \(F\) is Lipschitz in the second variable, that is, there exists a number \(L\) such that $$ |F(x, y)-F(x, z)| \leq L|y-z| \quad \text { for all } y, z \in J, x \in I . $$ Show that \(F\) is continuous as a function of two variables. Therefore, the hypotheses in the theorem could be made even weaker.
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