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Real Analysis: Math 4050-4060

Valerio De Angelis

Section 3.2 Limits

In the Calculus courses, we describe limits informally by often using the words “gets closer and closer”. Using that informal approach, we can describe the limit of a sequence as follows:
We say that the limit of a sequence \((a_n)\) is \(a\) if \(a_n\) gets closer and closer to \(a\) as \(n\) gets larger and larger.
But how exactly should we interpret “closer and closer”? And when we say that \(n\) gets large, how large is that?
We will now turn this informal concept into a completely precise one. We will give two definitions, one using almost only words, and the other using only symbols. Both provide some useful viewpoint. To clarify the words we use, we say that a number \(a\) is within \(\epsilon\) of another number \(b\) if their difference (in absolute value) is less than \(\epsilon\text{,}\) that is, if \(|a-b|\lt \epsilon\text{.}\)

Subsection 3.2.1 Definition of limit

We say that a sequence \((a_n)\) converges to a real number \(a\) if the following statement is true:
For every positive real number \(\epsilon\text{,}\) we can find some positive integer \(n_0\) with the property that \(a_n\) will be within \(\epsilon \) of \(a\) as long as \(n\) is at least as large as \(n_0\text{.}\)
The same statement using only symbols is as follows:
\begin{equation} (\forall \epsilon \gt 0)(\exists n_0\in {\mathbb N})(n\geq n_0 \Longrightarrow |a_n-a|\lt \epsilon)\tag{3.2.1} \end{equation}
It’s important to understand that both versions of the statement are saying the same thing. And a little thought will reveal that the statement is really the precise version of the informal one used in the Calculus courses.
If \((a_n)\) converges to \(a\) we write
\begin{equation*} \lim_{n\rightarrow \infty} a_n = a \end{equation*}
and we say that \(a\) is the limit of the sequence \((a_n)\). If there is no such \(a\text{,}\) we say that the sequence diverges.
We can say informally that proving that \(\displaystyle \lim_{n\rightarrow \infty} a_n=L\) boils down to finding a positive integer \(n_0\) (depending on \(\epsilon\)) that “works”, meaning that the inequality \(|a_n-L|\lt \epsilon\) will be true as soon as \(n\geq n_0\text{.}\) In fact we will often use this informal language to summarize steps in a proof. So if we are studying the limit \(\displaystyle \lim_{n\rightarrow \infty} a_n =a\text{,}\) we will say that \(n_0\) “works” for \((a_n)\) and \(\epsilon\) to mean that \(|a_n-a|\lt \epsilon\) whenever \(n\geq n_0\text{.}\)
The following lemma is often useful when proving results about limits.

Subsection 3.2.2 Examples

We now discuss a few examples to show how the definition of limit is used “in practice” to show that a sequence converges.

Example 3.2.2.

One of the simplest examples is the limit of the sequence \(1/n\text{.}\) We know that as \(n\) gets larger and larger, \(1/n\) will get smaller and smaller. So we want to prove that \(\displaystyle \lim_{n\rightarrow \infty}\frac{1}{n}=0\text{.}\) Given any \(\epsilon \gt 0\text{,}\) the inequality \(1/n \lt \epsilon \) is equivalent to \(n\gt 1/\epsilon\text{.}\) So, using the Archimedean principle, we can find some \(n_0\gt 1/\epsilon\text{,}\) and then if \(n\geq n_0\text{,}\) we have \(|1/n-0|=1/n \leq 1/n_0 \lt \epsilon\text{.}\) In this example, the integer \(n_0\) that “works” is any positive number \(n_0\geq 1/\epsilon\)
As another example on how to use the definition of limit, we will prove that the sequence \(\frac{2n}{n+1}\) has limit \(2\) as \(n\) gets large.

Example 3.2.3.

Let \(a_n=\dfrac{2n}{n+1}\text{,}\) \(n\geq 1\text{.}\) It is easy to see that this sequence converges and the limit must be \(2\text{:}\) if we divide numerator and denominator by \(n\text{,}\) we find
\begin{equation*} a_n=\frac{2}{1+1/n} \end{equation*}
and since \(1/n\) gets close to zero as \(n\) gets large, the fraction will get close to \(2\text{.}\) In fact, this is is how this result is derived in a Calculus course.
But in this course, we want to be more precise and rely only on the definition of limit. So we want to show that the difference between \(a_n\) and \(2\) gets smaller than any given positive number \(\epsilon\) when \(n\) is large enough. In symbols, we want to find some \(n_0\) such that
\begin{equation*} \left|\frac{2n}{n+1}-2\right|\lt \epsilon \mbox{ when } n \geq n_0. \end{equation*}
To solve the problem, we do some “reverse engineering” on the last inequality: we try to solve the inequality to see how large \(n\) needs to be. So we want to solve
\begin{equation*} \left|\frac{2n}{n+1}-2\right|\lt \epsilon \end{equation*}
or
\begin{equation*} \left|\frac{-2}{n+1}\right|\lt \epsilon \end{equation*}
or
\begin{equation*} \frac{2}{n+1}\lt \epsilon . \end{equation*}
So we need \(n+1\gt 2/\epsilon \text{,}\) or \(n\gt 2/\epsilon -1\text{.}\) This is telling us what we can choose for \(n_0\text{:}\) it will be enough to take any positive integer \(n_0\geq 2/\epsilon\text{.}\) Note that to simplify things we use \(2/\epsilon\) and not \(2/\epsilon -1\text{,}\) because any number larger than \(2/\epsilon\) will also be larger than \(2/\epsilon -1\text{.}\) At this point we can invoke the Archimedean principle: we know that there will be some positive integer \(n_0\) such that \(n_0\geq 2/\epsilon\text{.}\) This will be the integer that “works”.
We can consider the last paragraph as “scratch work” for the “official” proof: now that we know what \(n_0\) needs to be, we can do the proof. The next paragraph is the “official” proof.
Suppose that \(\epsilon\gt 0\) is given, and, using the Archimedean principle, find some positive integer \(n_0\) such that \(n_0\geq 2/\epsilon\text{.}\) We now need to show that if \(n\geq n_0\text{,}\) then \(|2n/(n+1)-2|\lt \epsilon\text{.}\) So suppose \(n\geq n_0\text{.}\) Then
\begin{equation*} \left|\frac{2n}{n+1}-2\right|=\left|\frac{-2}{n+1}\right|=\frac{2}{n+1}\lt \frac{2}{n}\leq \frac{2}{n_0}\leq \epsilon. \end{equation*}
And we are done.

Example 3.2.4.

Let \(a_n=\frac{3\sqrt{n}+1}{2\sqrt{n}-1}\text{.}\) It is easy to guess what the limit as \(n\rightarrow \infty\) is going to be: dividing numerator and denominator by \(\sqrt{n}\) we find
\begin{equation*} a_n=\frac{3+1/\sqrt{n}}{2-1/\sqrt{n}}. \end{equation*}
Since \(1/\sqrt{n}\) becomes smaller and smaller as \(n\) gets large, the limit must be \(3/2\text{.}\) But we want to prove it using the definition.
We calculate the absolute value of the difference \(a_n-3/2\text{,}\) with the aim to show that it becomes as small as we want. We find:
\begin{align*} \amp \left|\frac{3\sqrt{n}+1}{2\sqrt{n}-1}-\frac{3}{2}\right|=\left|\frac{2(3\sqrt{n}+1)}{2(2\sqrt{n}-1)}-\frac{3(2\sqrt{n}-1)}{2(2\sqrt{n}-1)}\right|\\ =\amp \left|\frac{6\sqrt{n}+2-6\sqrt{n}+3}{2(2\sqrt{n}-1)}\right|=\frac{5}{2(2\sqrt{n}-1)}. \end{align*}
Note that we have dropped the absolute value sign because the expression is clearly positive when \(n\geq 1\text{.}\) So we want this last expression to be less than a given positive number \(\epsilon\text{.}\) In other words we solve the inequality:
\begin{equation*} \frac{5}{2(2\sqrt{n}-1)}\lt \epsilon \end{equation*}
for \(n\text{:}\)
\begin{align*} \frac{2(2\sqrt{n}-1)}{5}\gt \amp\frac{1}{\epsilon}\\ 2\sqrt{n}-1\gt \amp\frac{5}{2\epsilon}\\ 2\sqrt{n}\gt \amp\frac{5}{2\epsilon}+1\\ \sqrt{n}\gt \amp\frac{1}{2}\left(\frac{5}{2\epsilon}+1\right)\\ n\gt \amp\frac{1}{4}\left(\frac{5}{2\epsilon}+1\right)^2 \end{align*}
The last line is telling us that if we take \(n_0\) to be any number greater than \(\frac{1}{4}\left(\frac{5}{2\epsilon}+1\right)^2\text{,}\) by retracing our steps we will find \(|a_n-3/2|\lt \epsilon\) (see the exercises). Note that, unlike in Example 3.2.3, we do not drop the \(1\) to simplify the \(n_0\) we need to \(\frac{1}{4}\left(\frac{5}{2\epsilon}\right)^2=\frac{25}{16\epsilon^2}\text{,}\) because doing so we would get a smaller number instead of a larger one.

Subsection 3.2.3 Convergent sequences are bounded

We are now ready to state the first important result on convergent sequences.
First find some \(n_0\) that “works” for \((a_n)\) and \(1\text{.}\) Then consider the two cases: \(n\lt n_0\) and \(n\geq n_0\text{.}\)

Subsection 3.2.4 Proving that a sequence diverges

Just like we sometimes want to rigorously prove that a sequence converges, other times we want to prove that it does not converge, or that it diverges. If the sequence is not bounded, than we can invoke Theorem 3.2.5 and conclude that the sequence cannot converge. So for example the sequence \(a_n=n\) cannot converge, because (by the Archimedean Principle) it is not bounded.
But if the sequence is bounded, than Theorem 3.2.5 is useless, and we need to do it by using the definition of convergence. This means that we need to consider its negation. This is best done by using the logical symbols. Saying that a sequence \((a_n)\) converges means that there is some real number \(a\) such that \(\displaystyle \lim_{n\rightarrow \infty} a_n=a\text{,}\) or, using symbols:
\begin{equation*} \left(\exists a\in {\mathbb R}\right)\left(\forall \epsilon\gt 0\right)\left(\exists n_0\in {\mathbb N}\right)\left(\forall n\geq n_0\right) \left(|a_n-a|\lt \epsilon\right). \end{equation*}
The negation of this statement is:
\begin{equation*} \left(\forall a\in {\mathbb R}\right)\left(\exists \epsilon\gt 0\right)\left(\forall n_0\in {\mathbb N}\right)\left(\exists n\geq n_0 \right)\left( |a_n-a|\geq \epsilon\right). \end{equation*}

Example 3.2.6.

We show that the sequence \(a_n=(-1)^n\) diverges. So we need to show that no matter what real number \(a\) we consider, the sequence does not converge to \(a\text{.}\) Since the sequence is always either \(1\) or \(-1\text{,}\) we consider three cases: \(a=1\text{,}\) \(a=-1\text{,}\) and \(|a|\neq 1\text{.}\) If \(a=1\text{,}\) we choose \(\epsilon =1\text{.}\) Then given any \(n_0\in {\mathbb N}\text{,}\) we can surely find an odd number \(n\geq n_0\text{,}\) and then \(|a_n-1|=|(-1)^n -1|=|-1-1|=2\gt 1\text{.}\) If \(a=-1\text{,}\) we also use \(\epsilon =1\text{,}\) and given any \(n_0\in {\mathbb N}\text{,}\) we can find some even \(n\geq n_0\) and then \(|a_n-(-1)|=|(-1)^n+1|=|1+1|=2\gt 1\text{.}\) If \(|a|\neq 1\text{,}\) we can choose \(\epsilon =\min\{|a-1|, |a+1|\}\) and then for any \(n\) we find that \(|a_n-a|=|(-1)^n -a|\) is either \(|a-1|\) or \(|a+1|\text{,}\) and so \(|a_n-a|\geq \epsilon\text{.}\)

Subsection 3.2.5 Properties of limits

Suppose we have a sequence \((a_n)\) converging to \(a\) and a sequence \((b_n)\) converging to \(b\text{.}\) We would expect that the sequence \((a_n+b_n)\) converges to \(a+b\text{,}\) and in fact that is true. But we need to prove it. In this section we address this and other similarly natural questions.
  1. Given \(\epsilon \gt 0\text{,}\) find \(n_0\) that “works” for \((a_n)\) and \(\epsilon/2\text{,}\) and find \(n_1\) that “works” for \((b_n)\) and \(\epsilon/2\text{.}\) Then use Lemma 3.2.1.
  2. Write \(a_nb_n-ab = b_n(a_n-a)+a(b_n-b)\) and use the triangle inequality. Use Theorem 3.2.5 to find some \(M\gt 0\) such that \(|b_n|\leq M\) for all \(n\text{.}\) Then find \(n_0\) that “works” for \((a_n)\) and \(\epsilon/(2M)\text{,}\) and find some \(n_1\) that “works” for \((b_n)\) and \(\epsilon/2(|a|+1)\text{.}\)
  3. Find \(n_0\) that “works” for \((a_n)\) and \(|a|/2\text{.}\) Then find \(n_1\) that “works” for \((a_n)\) and \(|a|^2\epsilon/2\text{.}\)
It should be clear that the limit of a sequence is not affected if we start the sequence at some point other than the beginning. In other words, if instead of the sequence \((a_n)=(a_1,a_2,a_3,\ldots)\) we consider the sequence \((a_m,a_{m+1},a_{m+2},\ldots)\text{,}\) where \(m\) is any fixed positive integer, then \(\displaystyle \lim_{n\rightarrow \infty}a_{m+n}=\displaystyle \lim_{n\rightarrow \infty}a_n\) (see the exercises).
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