Section 6.4 Quantifiers and rigorous limits
Quantifiers appear all over mathematics, and it is essential that you become comfortable reading, understanding and applying them. One particularly important application of quantifiers is to make the notions of limits rigorous. You have
51 encountered the idea of a
limit when you studied differential calculus; the idea that the value of a function gets “closer and closer” to a particular value as we take the argument of that function “closer and closer” to (say) zero. Quantifiers allow us to make “closer and closer” rigorous
52 . We will start with the idea of the limit of a sequence, and then move on to limits of functions. So we begin with the definition of a sequence.
Subsection 6.4.1 Convergence of sequences
Definition 6.4.1.
A sequence is an ordered list of real numbers. It is typically denoted
\begin{equation*}
(x_n)_{n \in \mathbb{N}} = (x_1, x_2, x_3, \dots)
\end{equation*}
The numbers \(x_1,x_2,\cdots\) are the terms of the sequence. You will also sometimes see alternate notation such as
\begin{equation*}
(x_n)_n, \qquad (x_n)_{n\geq 1} \qquad \text{or} \qquad (x_n)
\end{equation*}
In some texts you will also see a sequence denoted with braces, \(\{x_n\}\text{;}\) we will not use that notation to avoid confusion with set notation.
An alternate way to think of a sequence is as a function that takes natural numbers and maps them to real numbers. So, for example, the sequence
\(\left(\frac{1}{n}\right)_n\) is just the function
\(f:\mathbb{N} \to \mathbb{R}\) defined by
\(f(n)=1/n\text{.}\) We will come back to functions in
Chapter 10. We also note that one can generalise this definition to sequences of other types of numbers or objects, but we will focus on sequences of real numbers.
We will typically define a particular sequence either by giving the first few terms,
\begin{equation*}
(x_n) = (2,3,5,7,11,\dots)
\end{equation*}
or by giving a formula for the \(n^\mathrm{th}\) term in the sequence:
\begin{equation*}
(x_n) = \left(\dfrac{1}{n}\right)
\end{equation*}
ie, the sequence
\((1,1/2, 1/3, 1/4,\dots)\text{.}\) Just as was the case with defining sets, we must make sure that we give the reader enough information to understand our definition. In this way, giving a formula for the
\(n^\mathrm{th}\) term is typically preferred
53 .
Typically one can compute the first few terms of a sequence by hand, and then the next many terms by computer. However, we are very often interested in the behaviour of the terms of the sequence as \(n\) becomes very large. So, for example, consider the sequences
\begin{alignat*}{4}
(a_n) \amp = \left( \frac{1}{n}\right)_n \amp = \amp \left(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots \right)\\
(b_n) \amp = \left( (-1)^n \right)_n \amp = \amp \left(1, -1, 1, -1,\dots \right)
\end{alignat*}
Here are plots of the first few terms to see how they behave.
Notice that in the first sequence, as \(n\) gets larger and larger, the term \(a_n\) gets closer and closer to 0. In this case we will say that “\(a_n\) converges to \(0\)”. In the second case, the sequence simply bounces back and forth between \(+1\) and \(-1\) and does not appear to “settle” to any particular value. In this case we will say that “\(b_n\) diverges”. Let’s see how we can turn this intuitive (but imprecise) understanding of convergence and turn it into a rigorous, precise definition.
To start towards that definition, let us rephrase convergence as a sort of two player game as we did for some of the nested quantifier examples earlier in this chapter, and focus on the example of \((a_n) = \left(\frac{1}{n}\right)\text{.}\) In this game, Player 1 has to choose a small positive number, and then Player 2 has to work out how big does \(n\) have to be so that we can guarantee that the distance between \(a_n\) and \(0\) is smaller than Player 1’s number. So, for example,
We can verify this by computing the distance
54 between two numbers using the absolute value, and then noting that
\begin{equation*}
n \gt 100
\qquad \implies \qquad
\frac{1}{n} \lt \frac{1}{100} = 0.01
\qquad \implies \qquad
\left| \frac{1}{n} - 0 \right| \lt 0.01.
\end{equation*}
This is just one instance, and Player 1 could have chosen \(0.01\) or \(2^{-30}\text{.}\) Indeed, Player 1 can choose any arbitrarily small positive number \(\epsilon\text{,}\) and then Player 2 can always work out how big to make \(n\) so that the distance between \(a_n\) and \(0\) is guaranteed to be smaller than \(\epsilon\text{.}\)
Player 1 says “I have chosen \(\epsilon \gt 0\text{,}\) please make the distance between \(a_n\) and \(0\) smaller than \(\epsilon\text{.}\)”
Player 2 does some thinking and replies “Choose any \(n \gt \frac{1}{\epsilon}\) and then it will work.”
Again, we can verify this:
\begin{equation*}
n \gt \frac{1}{\epsilon}
\qquad \implies \qquad
\frac{1}{n} \lt \epsilon
\qquad \implies \qquad
\left| \frac{1}{n} - 0 \right| \lt \epsilon
\end{equation*}
That is, Player 2 can set some point in the sequence \(N = \frac{1}{\epsilon}\text{,}\) so that for every value of \(n \gt N\text{,}\) we can guarantee that \(|a_n - 0 | \lt \epsilon\text{.}\) We can rephrase this outcome as
No matter which \(\epsilon \gt 0\) that Player 1 chooses, Player 2 can always find some \(N\) so that if \(n \gt N\) then \(|a_n-0| \lt \epsilon\text{.}\)
If we now try to play the same game with the second sequence \((b_n) = \left( (-1)^n \right)\text{,}\) then we will quickly see that it does not work. Let us attempt to show that the sequence converges to \(1\) using the same game
Player 1 says “Make the distance between \(b_n\) and \(1\) smaller than \(0.01\text{.}\)”
Player 2 does some thinking and replies “Well, if \(n\) is even then this will be true, but it will fail for every single odd value of \(n\text{.}\) I cannot give you a guarantee.”
Player 2 cannot win. When \(n\) is even, \(b_n=1\) and so \(|b_n-1|=0 \lt 0.01\text{,}\) but when \(n\) is odd, \(b_n=-1\) and so \(|b_n-1|=2 \gt 0.01\text{.}\) So there is no way to make \(n\) sufficiently big, ie all \(n\) bigger than some \(N\text{,}\) so that \(|b_n - 1| \lt 0.01\text{.}\) We can rephrase this outcome as
There is a choice of \(\epsilon\) so that no matter which \(N\) Player 2 chooses, there will always be some \(n \gt N\) so that \(|b_n - 1 | \gt \epsilon\text{.}\)
This example will help us understand how to turn the intuitive idea of convergence into the rigorous definition.
Subsubsection 6.4.1.1 Quantifying towards a definition
To start towards our definition, we need to first describe all the objects involved, and then we explain that convergence means that the objects satisfy certain conditions. In this case, we will start with something like
Let \((x_n)_n\) be a sequence of real numbers, that is, \(x_n\in\mathbb{R}\text{,}\) for all \(n\in\mathbb{N}\text{.}\) Then we say that \((x_n)\) converges to a real number \(L\) when …
and now we need to explain those conditions. Let us start with our intuitive idea
Let \((x_n)_n\) be a sequence of real numbers, that is, \(x_n\in\mathbb{R}\text{,}\) for all \(n\in\mathbb{N}\text{.}\) Then we say that \((x_n)\) converges to a real number \(L\) when \(x_n\) gets closer and closer to \(L\) as \(n\) becomes larger and larger.
While this is quite descriptive, it is not actually all that precise. Let us rewrite it as follows:
Let \((x_n)_n\) be a sequence of real numbers, that is, \(x_n\in\mathbb{R}\text{,}\) for all \(n\in\mathbb{N}\text{.}\) Then we say that \((x_n)\) converges to a real number \(L\) when \(x_n\) moves arbitrarily close to \(L\) as \(n\) becomes larger and larger.
Now this means that in order for \(x_n\) to converge to \(L\text{,}\) it must be possible to bring the number \(x_n\) as close to \(L\) as we want by making \(n\) really big. This is exactly like the two player game we described above: Player 1 can make any choice of \(\epsilon\) and then Player 2 has to work out how large to make \(n\text{.}\) Let us write our definition again:
Let \((x_n)_n\) be a sequence of real numbers, that is, \(x_n\in\mathbb{R}\text{,}\) for all \(n\in\mathbb{N}\text{.}\) Then we say that \((x_n)\) converges to a real number \(L\) when no matter what positive \(\epsilon\) we choose, then for all \(n\) sufficiently large, we have \(|x_n - L | \lt \epsilon\text{.}\)
Now, in our game, Player 2 established just how big \(n\) needs so that we can guarantee the sequence terms are close enough to \(L\text{.}\) We denoted that threshold by \(N\text{,}\) and it will almost always depend on \(\epsilon\text{.}\) Of course, since Player 1 chooses \(\epsilon\) first, Player 2 can pick \(N\) with full knowledge of what Player 1 did. Time for another attempt at the definition:
Let \((x_n)_n\) be a sequence of real numbers, that is, \(x_n\in\mathbb{R}\text{,}\) for all \(n\in\mathbb{N}\text{.}\) Then we say that \((x_n)\) converges to a real number \(L\) when for all \(\epsilon \gt 0\), there is \(N \in \mathbb{N}\), so that for all natural numbers \(n \gt N\) we have \(|x_n - L | \lt \epsilon\text{.}\)
This is now pretty good. All that remains is to tweak the phrasing a little, clean it up and make it into a formal definition
55 .
Definition 6.4.2.
Let \((x_n)\) be a sequence of real numbers. We say that \((x_n)\) has a limit \(L\in\mathbb{R}\) when
\begin{equation*}
\forall\epsilon \gt 0, \exists N\in\mathbb N \st \forall n \in \mathbb{N}, (n\gt N) \implies (|x_n-L| \lt \epsilon).
\end{equation*}
In this case we say that the sequence converges to \(L\) and write
\begin{equation*}
x_n\to L \qquad \text{or} \qquad \lim\limits_{n\to \infty}x_n=L.
\end{equation*}
If the sequence doesn’t converge to any number \(L\text{,}\) we say that the sequence diverges.
Subsubsection 6.4.1.2 Some examples
Time to put our newly rigorised
56 understanding of limits to work.
Example 6.4.3.
Let \(c \in \mathbb{R}\text{.}\) Show that the constant sequence \((x_n)_{n\in\mathbb N}=(c)_{n\in\mathbb N}\) converges to \(c\text{.}\)
Scratchwork.
We know, at least intuitively, that the constant sequence must converge to that constant value its elements take. It makes sense. But, how can we prove it rigorously using the definition of sequence convergence.
Again, think of the two player game
57 . We need to show that no matter which
\(\epsilon \gt 0\) that Player 1 chooses, Player 2 can always find a threshold
\(N \in \mathbb{N}\) (which, in general will depend on
\(\epsilon\)), so that whenever
\(n \gt N\) we have
\(|x_n-c| \lt \epsilon.\)
Now assume that Player 1 picked an \(\epsilon>0\) and try to understand what Player 2 needs to show. They want to show that with the right choice of \(N\in\mathbb N\text{,}\) we can keep \(|x_n-c| \lt \epsilon\text{.}\) But since \(x_n\) constant \(x_n=c\text{,}\) we can simplify the inequality that needs to be satisfied:
\begin{equation*}
|x_n-c| \lt \epsilon \qquad \text{becomes} \qquad |c-c|=0 \lt \epsilon.
\end{equation*}
But since we know \(\epsilon \gt 0\text{,}\) this will be true independent of \(n\text{.}\) This, in turn, means that Player 2 is free to choose any \(N\in\mathbb N\text{.}\) Now Player 2 could be mysterious and pick any random value (eg \(N=98127\)) but let’s force Player 2 to be a bit nicer to the audience (ie to the reader) and get them to pick a sensible value like \(N=1\text{.}\) Then for all \(n \gt N=1\text{,}\) we have \(|x_n-c|=|c-c|=0 \lt \epsilon\) as required.
All that remains is to write this nice and tidy in a proof.
Proof.
Let \(\epsilon \gt 0\) be given and pick \(N=1\text{.}\) Then, for all \(n \gt N\text{,}\) we have
\begin{equation*}
|x_n - c| = |c-c| = 0 \lt \epsilon.
\end{equation*}
Therefore we conclude that \((x_n)\) converges to \(c\) as required.
Notice that the proof in our example does not have to explain how we came up with the choice of \(N\text{,}\) it merely has to show that it works. Our definition of convergence requires that “there exists \(N\)”, not that “there exists \(N\) with a nice explanation of how it was found”. This can sometimes make a nicely written proof seem much more clever than it really is. The reader can be left thinking “How on earth did they know how to choose that?”. But you should remember that the author of the proof probably did a lot of careful scratch work to work out how to make the proof nice and neat; that scratch work is not required for the proof to be valid. Of course, if the author knows their audience well, and knows they might need some help, then the author will hopefully leave some explanation in the text nearby.
Example 6.4.4.
Show that the sequence \((a_n)_{n\in\mathbb N}=(\frac 1n)_{n\in\mathbb N}\) converges to \(0\text{.}\)
Scratchwork.
This is precisely the example we did above to introduce the idea of convergence. We have even discussed how to think of proving the convergence as a two player game. So, let us dispense with the game analogy.
We are going to start by choosing some arbitrary \(\epsilon \gt 0\text{,}\) and then we need to find a threshold \(N\) so that for every \(n \gt N\text{,}\) we know that \(|a_n - 0 | \lt \epsilon\text{.}\) Now, our argument from the previous example won’t work here, since \(|a_n - 0| = \frac{1}{n}\) is no longer constant and varies with \(n\text{.}\) Let us try rewriting things a little to help us think.
We’ve chosen some arbitrary \(\epsilon \gt 0\text{,}\) and we need to show that \(|a_n-0| = \frac 1n \lt \epsilon\) when \(n\) is big enough, ie for \(n \gt N\text{.}\) But the inequality
\begin{equation*}
\frac{1}{n} \lt \epsilon \qquad \iff \qquad n \gt \frac{1}{\epsilon}
\end{equation*}
since both \(n, \epsilon\) are positive. So this means that given our \(\epsilon\text{,}\) we need to work out how to make sure that \(n \gt \frac{1}{\epsilon}\text{,}\) because then we know that \(\frac{1}{n} \lt \epsilon\text{.}\)
Well, if we set \(N = \frac{1}{\epsilon}\text{,}\) then when we require \(n \gt N\text{,}\) then we have that \(n \gt N \gt \epsilon\) and so \(\frac{1}{n} \lt \frac{1}{N} \lt \frac{1}{\epsilon}\) implying that \(\frac{1}{n} \lt \frac{1}{\epsilon}\text{.}\) The only hitch is that our definition requires that \(N\) be a natural number. But that is easy to fix, instead of setting \(N = \frac{1}{\epsilon}\text{,}\) we can just take \(N\) to be the next integer bigger than \(\frac{1}{\epsilon}\text{.}\) That is, we set \(N\) to be the ceiling of \(\frac{1}{\epsilon}\) which we write as \(N = \ceil{ \frac{1}{\epsilon} }\text{.}\)
Here, we should mention that even though any
\(N>\frac 1\epsilon\) would work for our arguments above and that we have infinitely many possible choices for
\(N\) it is important to either picking a specific
\(N\) as we did in this scratch work, or prove the existence of such an
\(N\) without giving it explicitly — using results like the Archimedean property
58 or similar. Simply saying that such number
should exist is not an adequate justification. With that caveat out of the way, let’s tidy this up and write the proof.
Proof.
Let \(\epsilon \gt 0\text{.}\) Then pick \(N=\ceil{ \frac{1}{\epsilon}}\text{,}\) so that \(N \geq \frac{1}{\epsilon}\text{.}\) Then for all \(n \gt N\text{,}\) we have that
\begin{equation*}
|a_n - 0 | = \frac{1}{n} \lt \frac{1}{N} = \epsilon
\end{equation*}
Therefore we see that \((a_n)\) converges to \(0\text{.}\)
Example 6.4.5.
Show that the sequence \((x_n) = \left(\dfrac{2n+4}{n+1}\right)\) converges to \(2\text{.}\)
Scratchwork.
Again, let’s start with the scratch work.
Our zeroth
59 step is to pick some arbitrary
\(\epsilon \gt 0\text{.}\) Now, as we saw in the previous example, we need to understand how to choose
\(n\) so that we can guarantee
\begin{equation*}
|x_n - L | = \left| \frac{2n+4}{n+1} - 2 \right| \lt \epsilon
\end{equation*}
To help us understand this, we should clean up the inequality and simplify the expression inside the absolute value. Simplifying shows us that we actually want
\begin{equation*}
\left|\frac{2n+4}{n+1} - 2\right| = \left|\frac{2n+4-(2n+2)}{n+1} \right|= \left|\frac{2}{n+1}\right| \lt \epsilon.
\end{equation*}
At this point we should reiterate — we have not proved this yet, this is just what we want to be true. We still need to work out how we choose \(n\) to make sure this is true.
We can even go a little further. Since we know \(n\) is a natural number, we know that \(\frac{2}{n+1} \gt 0\) and so we can write
\begin{equation*}
\left|\frac{2}{n+1}\right| = \frac{2}{n+1} \lt \epsilon.
\end{equation*}
This is a little easier to manipulate, and we can quickly isolate \(n\text{:}\)
\begin{equation*}
\frac{2}{n+1} \lt \epsilon
\qquad \iff \qquad
n+1 \gt \frac{2}{\epsilon}
\qquad \iff \qquad
n \gt \frac{2}{\epsilon}-1
\end{equation*}
So, this means that if we have \(n \gt \frac{2}{\epsilon}-1\text{,}\) then we know (moving back along our chain of reasoning) that \(|x_n-2| \lt \epsilon\text{.}\) So it makes sense for us to choose \(N=\ceil{ \frac{2}{\epsilon}} -1\text{.}\) Again, we make use of the ceiling function to ensure that \(N\) is an integer.
Oops! But be careful — what happens if
\(\epsilon=1000\text{?}\) Then we choose
\(N=0\) which is not a natural number. Thankfully this is easily fixed, let us just take
\(N\) to be a little bit larger. Indeed, we can set
\(N = \ceil{\frac{2}{\epsilon}}\) and everything works out nicely. More generally, in these types of proofs, once you have worked out an
\(N\text{,}\) one is free to make it
larger (you should ask yourself why). One could also argue that the choice
\(N = \ceil{ \frac{2}{\epsilon} }\) is a little neater
60 and does not change the proof very much.
Proof.
Let \(\epsilon \gt 0\text{,}\) set \(N = \ceil{(2/\epsilon)}\) and let \(n \gt N\text{.}\) By our choice of \(n\text{,}\) we know that
\begin{equation*}
n \gt N \gt \frac{2}{\epsilon}
\end{equation*}
From this we know that \(n+1 \gt n \gt \frac{2}{\epsilon}\) and so
\begin{equation*}
\frac{2}{n+1} \lt \epsilon
\end{equation*}
Hence
\begin{equation*}
|x_n-2|
= \left| \frac{2n+4}{n+1} - 2 \right|
= \left| \frac{2}{n+1} \right|
= \frac{2}{n+1}
\lt \epsilon,
\end{equation*}
as required.
Of course we also know that not every sequence converges — we have already seen an example of this above. Let us redo that example using our rigorous definition of convergence.
Example 6.4.6.
Show that the sequence \((b_n) = \left( (-1)^n\right)\) does not converge to \(1\text{.}\)
Scratchwork.
Even though we wish to show that the sequence \((b_n)\) does not converge to \(1\text{,}\) our starting point will still be the definition of convergence. Recall that the sequence \((b_n)\) converges to \(1\) when
\begin{equation*}
\forall\epsilon \gt 0, \exists N\in\mathbb N \st \forall n \in \mathbb{N} (n\gt N) \implies (|b_n-1| \lt \epsilon).
\end{equation*}
We want to show that this is false, and we do so by showing that the negation is true. The negation is
\begin{equation*}
\exists\epsilon \gt 0 \st \forall N\in\mathbb N, \exists n\in\mathbb{N} \st (n \gt N) \land (|(-1)^n-1|\geq\epsilon).
\end{equation*}
So, if we can show that this is true, then the original statement is false, and so the sequence \((b_n)\) does not converge to \(1\text{.}\)
Now, let’s try to understand this statement. It says that we need to show that there is an \(\epsilon \gt 0\) such that no matter what \(N\in\mathbb N\) we choose, there is always at least one \(n\gt N\) such that \(|(-1)^n-1|\) is greater than \(\epsilon\text{.}\) Now, notice that we don’t have to show this for all \(\epsilon\) (indeed we can’t), we just need to find one \(\epsilon\) that makes things work.
As we saw above, sequence \((b_n)\) alternates between \(-1\) and \(1\text{.}\)
For \(n\) even, \(b_n = (-1)^n = 1\) and so \(|b_n-1|=0 \lt \epsilon\text{.}\) So this is true for all \(\epsilon \gt 0\) and all even \(n\text{.}\)
On the other hand, for \(n\) odd, we have \(b_n = (-1)^n = -1\text{,}\) and so \(|b_n-1| = |-1-1| = 2\text{.}\) So, as long as we choose \(0 \lt \epsilon \lt 2\text{,}\) this case will fail for all odd \(n \in \mathbb{N}\text{.}\)
So to make the proof work, we can choose, say, \(\epsilon =1\) and then show that things go wrong for odd \(n\text{.}\) Time for the proof.
Proof.
We show that the sequence \((b_n)=\left((-1)^n\right)\) doesn’t converge to \(1\text{.}\) To this we show that
\begin{equation*}
\exists\epsilon \gt 0 \st \forall N\in\mathbb N, \exists n\in\mathbb{N}\st (n \gt N) \land (|(-1)^n-1|\geq\epsilon).
\end{equation*}
Let \(\epsilon=1\) and let \(N\) be any natural number. Now set \(n = 2N+1\text{,}\) so that \(b_n = (-1)^n=-1\text{.}\) Then
\begin{equation*}
|b_n -1 | = |-1-1| = 2 \gt \epsilon.
\end{equation*}
Therefore we conclude that the sequence does not converge to \(1\text{.}\)
Notice that we set \(n=2N+1\) in our proof and everything worked out nicely. We can actually choose any odd number larger than \(N\text{.}\) In fact, we could change the wording of the proof to say “Let \(n\) be any odd number larger than \(N\)” and it would be correct. But, since we can can make a simple explicit choice, we should do so.
In general, when we talk about the divergence of a sequence, we don’t say that the sequence does not converge to a given specific number
\(L\text{.}\) Rather, we typically want to prove
61 that it does not converge to
any number
\(L\text{.}\) This is a much stronger statement. Written as a quantified statement, it is
\begin{equation*}
\forall L\in\mathbb R, \exists\epsilon \gt 0 \st \forall N\in\mathbb N, \exists n\in\mathbb{N} \st (n\gt N) \land (|x_n-L|\geq\epsilon),
\end{equation*}
and says that for any number \(L\text{,}\) the sequence doesn’t converge to that number.
Let’s do an example where we show that a sequence actually diverges. That is, it does not converge to any number \(L \in \mathbb{R}\text{.}\)
Example 6.4.7.
Show that the sequence \((x_n)=(n)\) diverges.
Scratchwork.
First, let’s write down what we want to show:
\begin{equation*}
\forall L\in\mathbb R, \exists\epsilon \gt 0 \st \forall N\in\mathbb N, \exists n\in\mathbb{N} \st (n\gt N) \land (|n-L|\geq\epsilon).
\end{equation*}
Since we need to make this work for every possible \(L\text{,}\) we let \(L\) be an arbitrary real number. Now, what we want is to satisfy \(|n-L|\geq \epsilon\) for some \(\epsilon\) and some \(n\text{.}\) It can be a little intimidating to try to this for an arbitrary \(L\text{,}\) so perhaps it is better to think about a few different \(L\)-values.
When \(L=0\text{,}\) the we can simplify \(|n -L| = |n| = n\text{.}\) Now since, \(n \in \mathbb{N}\text{,}\) we know that \(n \geq 1\text{.}\) So if we pick \(\epsilon=1\text{,}\) then we will have \(|n-L| \geq \epsilon\) for all \(n \in \mathbb{N}\text{.}\) Now, it is easy to also enforce the requirement that \(n \gt N\text{,}\) no matter what \(N\) is chosen, just pick \(n = N+1\text{.}\)
Similarly, if we set \(L=-1\text{,}\) then we have \(|n-L|=|n-(-1)|=n+1\text{.}\) So again, we can pick \(\epsilon=1\) and then \(|n-L| \geq \epsilon\) for all \(n \in \mathbb{N}\text{.}\) This reasoning will actually work for any \(L \leq 0\text{.}\) Again, to also enforce the requirement that \(n \gt N\text{,}\) we can just pick \(n=N+1\text{.}\)
What about when, say, \(L=17\text{,}\) we will have \(|n-L| = |n-17|\text{.}\) Now, provided \(n \gt 17\text{,}\) this will be bigger than zero. In particular, if we set \(n \geq 18\text{,}\) then we will have \(|n-17| \gt 1\text{.}\) However, this is not quite right. We not only need that \(|n-17| \geq \epsilon\) but we also need that \(n \gt N\text{,}\) no matter what choice of \(N \in \mathbb{N}\text{.}\) Thankfully this is easily fixed, just choose \(n = \max\{17, N\}+1\text{.}\) Alternatively, since we know \(N \geq 1\text{,}\) we can choose \(n = N+17\text{.}\)
More generally, if \(L \gt 0\text{,}\) then we can choose \(n \geq \ceil{L}+1\text{,}\) and then \(|n-L| \geq 1\text{.}\) Just as in the previous point, we need to satisfy \(n \gt N\text{,}\) so pick \(n = \max\left\{\ceil{L},N \right\}+1\text{.}\) Notice that a similar choice works when \(L \leq 0\text{,}\) just take \(n = \max\left\{\ceil{|L|},N \right\}+1\text{.}\)
We are now ready to write the proof.
Proof.
Let \(L\) be an arbitrary real number and set \(\epsilon = 1\text{.}\) Then for any \(N \in \mathbb{N}\text{,}\) set \(n = \max\{N, \ceil{|L|} \}+1\text{.}\) Then this gives
\begin{equation*}
n \gt N \qquad \text{ and } \qquad |n-L| \gt 1=\epsilon
\end{equation*}
Therefore the sequence \((x_n)=(n)\) diverges.
Notice how short the proof is compared to the scratch work. This is not unusual. A nice neat proof can hide a lot of work.
Subsection 6.4.2 The limit of a function
Note that in this section of the text we restrict ourselves to real-valued functions. That is, functions that take a real number as input and return a real number as output, just like those you worked with in Calculus courses. We do look at more general functions in
Chapter 10, but not their limits.
We define the limits of functions in very much the same way as the limits of sequences. The definition is more general, as now we can talk of the limit of a function as its argument approaches more general points, while for a sequence, we only talked of its behaviour as \(n \to \infty\text{.}\) Let’s give the definition and then we’ll explain it.
Definition 6.4.8.
Let \(a,L \in \mathbb{R}\) and let \(f\) be a real-valued function. We say that the limit of \(f\) as \(x\) approaches \(a\) is \(L\) when
\begin{equation*}
\forall\epsilon \gt 0, \exists \delta \gt 0 \st \left(0\lt |x-a| \lt \delta\right)\implies \left(|f(x)-L| \lt \epsilon \right).
\end{equation*}
In this case we write
\begin{equation*}
\lim\limits_{x\to a}f(x)=L \qquad \text{or sometimes}\qquad f(x)\xrightarrow[x\to a]{} L
\end{equation*}
and say that \(f\) converges to \(L\) as \(x\) approaches \(a\text{.}\) We also sometimes say the limit of \(f\) as \(x\) goes to \(a\) is \(L\text{,}\) which we denote by
\begin{equation*}
f(x) \to L \text{ as } x \to a.
\end{equation*}
If \(f\) does not converge to any finite limit \(L\) as \(x\) approaches \(a\text{,}\) then we say that \(f\) diverges as \(x\) approaches \(a\text{.}\)
This definition may look
62 more complicated than the equivalent definition for the convergence of a sequence,
Definition 6.4.2. But if we do some reverse engineering (much as we did for sequence convergence above, but in reverse) then we can read the definition as
For all positive \(\epsilon\text{,}\) there is some positive \(\delta\) so that if the distance between \(x\) and \(a\) is less than delta (but not zero), then the distance between \(f(x)\) and \(L\) is less than \(\epsilon\text{.}\)
We can rephrase these quantifiers as little:
No matter which positive \(\epsilon\text{,}\) we can always find some positive \(\delta\) so that if the distance between \(x\) and \(a\) is less than delta (but not zero), then the distance between \(f(x)\) and \(L\) is less than \(\epsilon\text{.}\)
This is telling us that if we need to make the distance between the function and its limit very small, we can always find some \(\delta\) so that we just need to make the distance \(|x-a| \lt \delta\text{.}\) That is
So we reach
This is probably, more or less, the working definition of a limit of a function you used in your first Calculus course. This gives reasonable intuition, but the power of quantifiers is to make everything precise and eliminate misunderstandings.
Let’s put this definiton to work by considering the limit of a simple function.
Example 6.4.10.
Show that for any \(\ds a\in \mathbb R, \lim_{x\to a}x=a\text{.}\)
Scratchwork.
In this example we want to show that the function \(f(x)=x\) converges to the limit \(a\) as \(x\) goes to \(a\text{.}\) Even though this feels more like a tautology than an example, it is a good exercise in applying the limit definition.
To prove that \(\lim_{x\to a}x = a\) the definition tells us that we need to show that
\begin{equation*}
\forall\epsilon \gt 0, \exists \delta\st 0 \st \left(0\lt |x-a|\lt \delta\right)\implies \left(|f(x)-L| \lt \epsilon \right).
\end{equation*}
Now, this simplifies immediately since we have \(f(x)=x\) and \(L=a\text{:}\)
\begin{equation*}
\forall\epsilon \gt 0, \exists \delta\st 0 \st \left(0\lt |x-a|\lt \delta\right)\implies \left(|x-a| \lt \epsilon \right).
\end{equation*}
So, given an arbitrary \(\epsilon>0\text{,}\) we need a \(\delta>0\) such that
\begin{equation*}
0 \lt |x-a| \lt \delta \implies |x-a| \lt \epsilon.
\end{equation*}
That is, whatever positive
\(\delta\) we pick, whenever
\(0 \lt |x-a| \lt \delta\text{,}\) it implies that
\(|x-a| \lt \epsilon\text{.}\) A good
63 choice for
\(\delta\) is simply
\(\delta = \epsilon\text{.}\)
Now, let’s write this in a proof.
Proof.
Suppose \(\epsilon\) is any positive real number. Then pick \(\delta = \epsilon\text{.}\) Then whenever \(|x-a| \lt \delta\text{,}\) then we know that \(|f(x)-a| = |x-a| \lt \delta =\epsilon\) as required.
That one is arguably a little too simple. Here is a slightly more complicated one.
Example 6.4.11.
Let \(a \in \mathbb{R}\text{.}\) Prove that
\begin{equation*}
\lim_{x\to a} 3x+5 = 3a+5.
\end{equation*}
Scratchwork.
Again, our starting point is to look at the definition. We need to prove that
\begin{equation*}
\forall\epsilon \gt 0, \exists \delta \gt 0 \st (0\lt |x-a| \lt \delta)\implies (|(3x+5)-(3a+5)| \lt \epsilon).
\end{equation*}
Before we go much further, we should clean this up a little. We can simplify that last inequality. That is \(|(3x+5)-(3a+5)| = |3x-3a| = 3|x-a|\text{,}\) so
\begin{equation*}
\forall\epsilon \gt 0, \exists \delta \gt 0 \st (0\lt |x-a| \lt \delta)\implies (3|x-a| \lt \epsilon).
\end{equation*}
Now this is looking pretty similar to the previous example. Given any \(\epsilon\text{,}\) we need to pick \(\delta\text{,}\) so that when \(0 \lt |x-a| \lt \delta\text{,}\) we guarantee that \(3|x-a| \lt \epsilon\text{.}\) If we rearrange this last inequality, we want
\begin{equation*}
|x-a| \lt \frac{\epsilon}{3}.
\end{equation*}
And thus we pick \(\delta = \frac{\epsilon}{3}\text{.}\)
Alternatively, if we assume that we have \(0 \lt |x-a| \lt \delta\text{,}\) then multiplying everything by \(3\) gives:
\begin{equation*}
0 \lt 3|x-a| \lt 3\delta
\end{equation*}
and thus we need \(3\delta \leq \epsilon\text{.}\) So again, we reach the neat choice of \(\delta = \frac{\epsilon}{3}\text{.}\) Time for the proof.
Proof.
Let \(\epsilon\) be any positive real number. Then pick \(\delta = \frac{\epsilon}{3}\text{.}\) Then as long as \(|x-a| \lt \delta\text{,}\) we have that
\begin{equation*}
|(3x+5)-(3a+5)| = |3x-3a| = 3|x-a| \lt 3\delta = \epsilon.
\end{equation*}
And thus \((3x+5)\) converges to \(3a+5\) as \(x\) approaches \(a\text{.}\)
As you can see, the proof of the statement is very short, clean, and doesn’t omit any necessary information. And, as was the case with our proofs above, we don’t explain to the reader how we come up with the choice of \(\delta = \epsilon/3\text{,}\) we just have to prove that it works.
Let’s ratchet up the difficultly a little.
Example 6.4.12.
Show that \(\ds \lim_{x\to2} \left(\dfrac{1}{x} \right)= \dfrac{1}{2}\text{.}\)
Scratchwork.
Just like in our previous example(s) we start with the definition of convergence and adapt it to the problem at hand. We need to show that
\begin{equation*}
\forall\epsilon \gt 0, \exists \delta \gt 0 \st (0\lt |x-2|\lt\delta)\implies \left(\left|\frac 1x-\frac 12\right|\lt \epsilon\right).
\end{equation*}
Again, we can clean up and simplify the final inequality, since
\begin{equation*}
\left|\frac{1}{x}-\frac{1}{2}\right| = \left|\frac{2-x}{2x} \right| = \frac{|x-2|}{2|x|}.
\end{equation*}
Thus we need to show that
\begin{equation*}
\forall\epsilon \gt 0, \exists \delta \gt 0 \st (0\lt |x-2|\lt\delta)\implies \left( \frac{|2-x|}{2|x|} \lt \epsilon\right).
\end{equation*}
Now that we know what we need to show, let \(\epsilon>0\) be arbitrary. Then, we want a \(\delta \gt 0\) such that if we assume \(0 \lt |x-2| \lt \delta\text{,}\) we can guarantee that \(\frac{|x-2|}{2|x|} \lt \epsilon\text{.}\) Well, if we know that \(|x-2| \lt \delta\text{,}\) then we can write
\begin{equation*}
\frac{|x-2|}{2|x|} \lt \frac{\delta}{2|x|} \lt \epsilon
\end{equation*}
and so we need
\begin{equation*}
\delta \lt 2|x| \epsilon.
\end{equation*}
This is not quite enough — our choice of \(\delta\) should not depend on \(x\text{.}\) We need some bound on \(x\text{.}\)
Recall our intuitive idea of the limit, as \(x\) gets very close to \(2\text{,}\) the function \(\frac{1}{x}\) gets very close to \(\frac{1}{2}\text{.}\) We don’t really care what happens when \(x\) is a long way from \(2\text{,}\) and only on what happens when \(x\) is very close to \(2\text{.}\) Thus we should be able to focus on the region around \(x=2\text{,}\) say, \(1 \lt x \lt 3\text{,}\) or equivalently, \(|x-2| \lt 1\text{.}\)
How
64 does this help us? Well, if we know that
\(1 \lt x \lt 3\text{,}\) then we know
65 that
\(|x|\gt 1\text{,}\) and so
\begin{equation*}
\frac{|x-2|}{2|x|} \lt \frac{\delta}{2|x|} \lt \frac{\delta}{2} \leq \epsilon
\end{equation*}
And thus we need to ensure that
\begin{equation*}
\delta \leq 2 \epsilon.
\end{equation*}
At this point it seems that we can choose any \(\delta \leq 2 \epsilon\text{,}\) but this is not quite right. Say, we chose a large value of \(\epsilon\text{,}\) like \(\epsilon=3\text{,}\) and so we could pick \(\delta = 2\epsilon=6\text{.}\) With that choice of \(\epsilon\) and \(\delta\text{,}\) the implication at the heart of the definition of convergence becomes
\begin{equation*}
(|x-2| \lt 6) \implies \left(\left|\frac{1}{x}-\frac{1}{2}\right| \lt 3\right).
\end{equation*}
Unfortunately this is false. We could take, say \(x=\frac{1}{10}=0.1\text{,}\) and then the hypothesis is true, since \(|x-2|=1.9 \lt 6\text{,}\) but the conclusion is false since \(\left|\frac{1}{x}-\frac{1}{2}\right| = |10-0.5| = 9.5 \gt 3\text{.}\)
What went wrong? Remember to make our bound on \(|x|\) we required that \(|x-2|\lt 1\text{.}\) This is the same as requiring that \(\delta\leq 1\text{.}\) So we have actually imposed two requirements on \(\delta\text{.}\) We need both \(\delta \leq 1\) and \(\delta \leq 2\epsilon\text{.}\) To enforce both of these we can pick
\begin{equation*}
\delta = \min\{1, 2\epsilon \}.
\end{equation*}
Now we can finally write up the proof.
Proof.
Let \(\epsilon \gt 0\) and set \(\delta = \min\{1,2\epsilon\}\text{.}\) Now assume that \(0 \lt |x-2| \lt \delta.\) Since \(\delta \leq 1\text{,}\) we know that \(|x-2| \lt 1\) and so \(1 \lt |x| \lt 3\) and thus \(2|x| \gt 2\text{.}\)
Then
\begin{equation*}
\left| \frac{1}{x} - \frac{1}{2} \right| = \frac{|x-2|}{2|x|} \lt \frac{|x-2|}{2} \lt \frac{\delta}{2}.
\end{equation*}
Since \(\delta \lt 2\epsilon\text{,}\) we know that
\begin{equation*}
\left| \frac{1}{x} - \frac{1}{2} \right| \lt \epsilon
\end{equation*}
and so \(\frac{1}{x} \to \frac{1}{2}\) as \(x \to 2\text{.}\)
At least we think you should have. The authors are making an assumption about your mathematical education here, and we appologise if you have, in fact, not encountered limits before now.
We recommend looking up the history of infinitesimals and fluxions which predate the rigorous definition of limits we give here. The calculus of Newton and Leibniz used infinitesimals to understand limits and derivatives. These ideas were attacked by Berkeley in his 1734 book “The Analyst” in which he refers to infinitesimals as “the ghosts of departed quantities”. The rigorous definition of limits, and so a rigorous foundation for calculus, was given almost 150 years later by Cauchy, Bolzano and Weierstrass. There is much of interest here for a motivated reader with a good search engine. We also recommend a little digression into surreal numbers, hyperreal numbers and nonstandard analysis — things can get pretty weird.
The reader should assume that the sequence
\((2,3,5,7,11,\dots)\) is just the prime numbers, but a little digging in the
Online Encyclopedia of Integer Sequences shows a few other possibilities (some reasonable and some weird) including “partitions” (the number of ways of writing a given integer as a sum of smaller positive integers), “additive primes” (the sum of the digits is also prime), “absolute primes” (every permutation of the digits is also a prime) and lengths of “Farey sequences” (the reader should search-engine this one). Of course, if we are doing a good job as an author then we will have provided the reader with enough context to determine the sequence correctly.
The absolute value is an example of a metric or distance function. You already know other examples of metrics — for example, the distance between points \((x,y)\) and \((z,w)\) on the Cartesian plane is given by \(d = \sqrt{(x-z)^2 + (y-w)^2}\text{;}\) this is called the Euclidean metric. Another way to compute distances on the Cartesian plane is the taxicab metric \(d = |x-z|+|y-w|\text{.}\) A search-engine will guide you to more on this topic.
This is, more or less, the definition given by Bolzano in 1816.
The authors were quite sure that “rigorised” is not a real word and were surprised to find it in the dictionary.
We’ll dispense with this analogy soon.
This says, roughly, that given any two positive real numbers, \(x,y\text{,}\) we can always find an integer \(n\text{,}\) so that \(nx \gt y\text{.}\) This appears in a more geometric guise in Book V of Euclid’s Elements, and Archimedes attributes it to Eudoxus.
This is barely a step at all really. But we do need \(\epsilon\text{.}\)
Mathematicians generally find “neatness” to be desirable in a proof. Of course, one should not make things so neat that the logic is obscured.
There are exceptions to this. For example, one easy way to show that a series diverges is to show that summands do not converge to zero.
In fairness, it is a little more complicated. But it is not that much more complicated.
There are an infinite number of possible correct choices of \(\delta\text{.}\) Indeed, \(\delta = \epsilon/n\) for any \(n \in \mathbb{N}\) works. But the choice of \(\delta = \epsilon\) is good because it works, while being neat and simple.
Should one ask rhetorical questions in a textbook?
Be careful with the inequalities here.
