https://en.wikipedia.org/wiki/Subsequential_limit
The first online source I found was from wikipedia and I decided to pick wiki because I feel like it was the best definition that I could find and the easiest to understand. I also like how with in the wiki page you can click on limit superior and limit inferior with in the definition and it will bring you to a new wiki page giving a definition and examples on inferior and superior.
http://mathonline.wikidot.com/limit-superior-and-limit-inferior
This online source does a great job explaining what a limits superior and inferior are and explains it by not just using words, but pictures as well. This source likes to use the number line to explain the inferior and superior and this source also gives out many definitions and examples.
Four examples:
http://math.stackexchange.com/questions/280260/what-are-some-examples-of-sequences-that-have-multiple-limit-points
https://math.berkeley.edu/~sagrawal/su15_math104/lec12_limsup.pdf
This second link has 4 exercises that could help become more familiar with the inferior and superior of limits.
2Y: Subsequential Limit Sets
By Samantha Johannes (Fall 2017)
The previous section 2X: Sequences and Subsequences provided an insight into sequences and subsequences. Now we’re going to jump into the limits of subsequences, subsequential limit sets, limit inferior and limit superior.
You already know the Bolzano-Weierstrass Theorem, which states that if it is a bounded sequence [latex]s_{n}[/latex], then there exists a convergent subsequence [latex]s_{n_{k}}[/latex]. We can utilize this theorem to draw some conclusions about limits and convergence. If we know that a bounded sequence has a convergent subsequence, then we know that the subsequence has a defined real number limit. This is because of the definition of convergence, which you have previously learned. This limit becomes the key in determining the bounds of the function itself. Although the Bolzano-Weierstrass Theorem merely states a convergent subsequence exists if a sequence is bounded, there are cases where there is more than one convergent subsequence. In order to understand the connection between the limits of convergence subsequences and limit sets, let’s look at a sequence that we learned in calculus.
Example:[latex]s_{n} = (-1)^n[/latex]
We know that this sequence is divergent. The limit of the function does not exist. However, it is bounded by the values -1 and 1, making it a bounded sequence. The[latex]n^{th}[/latex]term in the sequence determines the pattern of the sequence. If[latex]n[/latex]is even, the sequence equals 1. If[latex]n[/latex]is odd, the sequence equals -1.
These two limits make up the set {[latex]-1, 1[/latex]}, which contains the limits of the convergent subsequences of our original sequence. Because it contains the limits of the subsequences of [latex](-1)^n[/latex]. We can describe “if [latex]n[/latex] is even, the sequence equals 1″ as the subsequence [latex]s_{n_{k}}=(-1)^{2k}[/latex] with the [latex]lim_{n\to\infty}s_{n_{2k}}=1[/latex]. The statement “if [latex]n[/latex] is odd, the sequence equals -1” is converted to the subsequence [latex]s_{n_{2k+1}}=(-1)^{2k+1}[/latex] with the [latex]lim_{n\to\infty}s_{n_{2k+1}}=-1[/latex].
The subsequential limit set, denoted by[latex]S'[/latex], of a sequence[latex]s_{n}[/latex]is the set of all limits of subsequences of[latex]s_{n}[/latex]:
[latex]S'[/latex]={t[latex]\in[/latex][latex]\mathbb{R}[/latex] there exists a subsequence[latex]s_{n_{k}}[/latex] such that [latex]lim_{n\to\infty}[/latex][latex]s_{n_{k}}[/latex]=t}
This is also referred to as the limit set [latex]S'[/latex], which is the set of the real number limits of the convergent subsequences ([latex]s_{n_{k}}[/latex]) of[latex]s_{n}[/latex]:
[latex]S'[/latex]={L[latex]\in[/latex][latex]\mathbb{R}[/latex]:[latex]\exists[/latex]{[latex]n_{k}[/latex]},[latex]lim_{k\to\infty}s_{n_{k}}=L[/latex]}
As you can see, these two definitions can be used interchangeably. They both state that a limit set includes the limits of the subsequences of the given sequence. In order to understand where the limit set comes from, let’s look at a few more examples.
Example: [latex]s_{n}=cos(\frac{n\pi}{3})[/latex]
At first glance, it appears that finding the subsequential limit set may be difficult because of the inclusion of cosine. However, we can begin calculating the values for the sequence to try and establish a pattern.
When [latex]n=1[/latex],[latex]s_{n}=cos(\frac{1\pi}{3})=cos(\frac{\pi}{3})= \frac{1}{2}[/latex].
When [latex]n=2[/latex],[latex]s_{n}=cos(\frac{2\pi}{3})=-\frac{1}{2}[/latex].
When [latex]n=3[/latex],[latex]s_{n}=cos(\frac{3\pi}{3})=cos({\pi})=-1[/latex].
When [latex]n=4[/latex],[latex]s_{n}=cos(\frac{4\pi}{3})=-\frac{1}{2}[/latex].
When [latex]n=5[/latex],[latex]s_{n}=cos(\frac{5\pi}{3})= \frac{1}{2}[/latex].
When [latex]n=6[/latex],[latex]s_{n}=cos(\frac{6\pi}{3})=cos({2\pi})=1[/latex].
When [latex]n=7[/latex],[latex]s_{n}=cos(\frac{7\pi}{3})= \frac{1}{2}[/latex].
When [latex]n=8[/latex],[latex]s_{n}=cos(\frac{8\pi}{3})=-\frac{1}{2}[/latex].
When [latex]n=9[/latex],[latex]s_{n}=cos(\frac{9\pi}{3})=cos (3\pi)= cos(\pi)=-1[/latex].
After calculating the first nine values of the sequence, we notice that certain limits are repeated on the interval[latex][0,2\pi][/latex]. After the first six terms, the limits of the terms repeat again. Therefore, we can focus on the first six term values to pull subsequence limits for our limit set.
The first six limit values within the interval[latex][0,2\pi][/latex]for[latex]n[/latex] from the first to sixth term are[latex]\frac{1}{2},-\frac{1}{2},-1,-\frac{1}{2},\frac{1}{2}[/latex]and[latex]-1[/latex]. Since some values are repeated in the first six terms, our subsequential limit set would be [latex]S'[/latex]={[latex]-1,-\frac{1}{2},\frac{1}{2},1[/latex]}.
Example: [latex]s_{n}=\frac{1}{n}[/latex]
What is the subsequential limit set of this sequence? Let’s observe the subsequences and their limits.
When [latex]n=1[/latex],[latex]s_{n}=\frac{1}{1}=1[/latex]. Therefore, the limit of this subsequence is 1.
When [latex]n=2[/latex],[latex]s_{n}=\frac{1}{2}[/latex], so the limit would also be [latex]\frac{1}{2}[/latex].
When [latex]n=3[/latex],[latex]s_{n}=\frac{1}{3}[/latex], making the limit [latex]\frac{1}{3}[/latex].
We could keep going here, but I think you see the pattern emerging regarding the limit of the subsequences. While the patterns does not relate to the odd or even values of n, it is associated with the value of n itself.
We can define the limit of the subsequence as[latex]lim_{n\to\infty}s_{n_{k}}=\frac{1}{k}[/latex]. Therefore, the subsequential limit set would be[latex]S'[/latex]={[latex]1,\frac{1}{2},\frac{1}{3},...,\frac{1}{k}[/latex]}.
Okay, now that we’ve reviewed some examples regarding subsequential limit sets, we can now introduce the concepts of what the limit superior and limit inferior are.
The limit superior of a sequence [latex]x_{n}[/latex], denoted as[latex]lim_{n\to\infty}sup x_{n}[/latex], is the infimum of all numbers[latex]\beta[/latex]with the property that:
There is an integer[latex]N[/latex]so that[latex]x_{n}<\beta[/latex]for all[latex]n\geq N[/latex].
The limit inferior of a sequence [latex]x_{n}[/latex], denoted as[latex]lim_{n\to\infty}inf x_{n}[/latex], is the supremum of all numbers[latex]\alpha[/latex]with the property that:
There is an integer[latex]N[/latex]so that[latex]\alpha < x_{n}[/latex]for all[latex]n\geq N[/latex].
By these two definitions,[latex]-\infty[/latex] and [latex]\infty[/latex] are not allowed to be the values for limit inferior or limit superior.
Let’s quickly review the definitions and descriptions for both supremum and infimum before exploring the importance of these two values.
The supremum for a set of real numbers that is bounded above and nonempty is the least of all upper bounds of E, denoted by[latex]M[/latex]. We write[latex]M=supE[/latex].
The infimum for a set of real numbers that is bounded below and nonempty is the greatest of all lower bounds of E, denoted by[latex]m[/latex]. We write[latex]m=infE[/latex].
The incorporation of the supremum and infimum into the definitions of limit superior and inferior may be confusing at first, especially since the infimum is linked to limit superior while the supremum is linked to limit inferior. Let’s try to discuss the concepts in our own terms to explain the distinction.
For the limit superior, the infimum is applied within the definition because you are choosing the greatest or largest value in the (subsequential) limit set. Don’t get caught up in the technical information involved in infimum, but focus on it meaning the greatest value. In more basic terms (without the inclusion of the infimum),[latex]lim_{n\to\infty}sup s_{n}=sup S'[/latex]. So you want to choose the largest value in the set.
Regarding the limit inferior, the supremum is applied within the definition because you are choosing the least or smallest value in the (subsequential) limit set. For the supremum, focus on it meaning the least value. In more basic terms,[latex]lim_{n\to\infty}inf s_{n}=inf S'[/latex]. So you want to choose the smallest value in the set.
If [latex]x_{n}[/latex] is a sequence comprised of real numbers, then [latex]lim_{n\to\infty}inf x_{n} \leq lim_{n\to\infty}sup x_{n}[/latex]. The proof of this theorem can be found in our textbook reading if you want a condensed and completed proof clearly outlining how to reach those findings.
If a sequence is bounded, the limit superior and inferior always exist. However, if a sequence is unbounded, the limit superior equals [latex]\infty[/latex] while the limit interior equals [latex]-\infty[/latex] if the sequence is unbounded below.
One theorem in our reading states: Let [latex]s_{n}[/latex] be a sequence of real numbers. Then [latex]s_{n}[/latex] is convergent with[latex]lim_{n\to\infty}s_{n}=L[/latex]if and only if[latex]lim_{n\to\infty}inf s_{n}=lim_{n\to\infty}sup=L[/latex]. Notice how it is an if and only if statement, meaning that it is true in both directions. Both the conditional and converse are true, so the if and only if section stands.
Now that we know about limit superior and inferior, let’s go back to the previously listed examples and find the specific limit superior and inferior values from the subsequential limit set.
Example:[latex]s_{n} = (-1)^n[/latex]
For this sequence, we have established that the limit set is [latex]S'=[/latex]{[latex]-1,1[/latex]} because of the pattern for the odd and even [latex]n^{th}[/latex] terms. Now we must determine what the limit superior and inferior are. We know they exist because of the previous part stating that a bounded sequence has a limit superior and inferior.
Let’s look at the limit superior first. Using the more basic terms of its definition, the limit superior is the greatest value in the limit set. We can see that this value is[latex]1[/latex], making the[latex]lim_{n\to\infty}sup (-1)^{n}=1[/latex].
Using the similar logic in reverse for the limit inferior, we know that it will be the lowest value in the limit set. That value is[latex]-1[/latex], making [latex]lim_{n\to\infty}inf (-1)^{n}=-1[/latex].
Example: [latex]s_{n}=cos(\frac{n\pi}{3})[/latex]
We found the subsequential limit set to be[latex]S'[/latex]={[latex]-1,-\frac{1}{2},\frac{1}{2},1[/latex]}.
Starting with the limit superior,[latex]lim_{n\to\infty}sup cos(\frac{n\pi}{3})=1[/latex]because [latex]1[/latex] is the greatest value in the set.
Moving onto the limit inferior,[latex]lim_{n\to\infty}inf cos(\frac{n\pi}{3})=-1[/latex]because [latex]-1[/latex] is the least value in the limit set.
You will come to find that for sequences involving [latex]cos[/latex]or[latex]sin[/latex], the limit superior and limit inferior will typically be [latex]1[/latex]and[latex]-1[/latex] (respectively) since both of those sequences are bounded by the same numbers which are the minima and maxima for the sequences. The functions themselves for cosine and sine are bounded, making the sequences bounded. However, if one of these two sequences is altered or modified with a number added or subtracted to create a new sequence, the limit superior and inferior change. One hint is to subtract one and add one to the number joined to the cosine or sine sequence. One of the documents listed below provides an example this scenario.
Example: [latex]s_{n}=\frac{1}{n}[/latex]
The subsequential limit set for this sequence is[latex]S'[/latex]={[latex]1,\frac{1}{2},\frac{1}{3},...,\frac{1}{k}[/latex]}.
The[latex]lim_{n\to\infty}sup \frac{1}{n}=1[/latex]because the first term equals [latex]1[/latex] and starts working its way lower and lower. All the fractions after [latex]1[/latex]are less than the first term.
Regarding the limit inferior, it is not directly included in the subsequential limit set. However, we do know that fractions keep getting smaller and smaller, getting closer and closer to zero. Now, we all know this terminology is not ideal when explaining the pattern of terms. So we state instead that[latex]lim_{n\to\infty}inf \frac{1}{n}=0[/latex]. This is also true because, according to the Archimedean principle, the natural numbers are greater than zero and unbounded. Therefore, the fraction can get arbitrarily close to zero but never equal zero since [latex]\frac{1}{0}[/latex] is undefined.
Here are some helpful documents, readings and examples to look at regarding sequential limit sets, limit superior and limit inferior:
http://www.math.ust.hk/~makyli/301_2008-9Fa/TutNt_05.pdf
http://cecas.clemson.edu/~petersj/Courses/M453/Lectures/L11-LiminfLimsupSeq.pdf
http://www.trillia.com/dA/zakon-analysisI-us-one.pdf
Resources/Open Math Education Links:
https://aimath.org/textbooks/approved-textbooks/thomson-bruckner-bruckner/ Elementary Real Analysis http://classicalrealanalysis.info/com/documents/TBB-AllChapters-Landscape.pdf
http://matthematics.com/ura/section-bolzano.html