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Section 5.1 Sequences and the definition of limit

How does a computer compute \(\sqrt{51}\text{?}\) A computer can't possibly memorise all possible square roots, so it must compute it from scratch. Moreover, it can't possibly compute the entire decimal expansion (next week we will see that, because \(\sqrt{51}\) is irrational, the decimal expansion never repeats), so it can only come up with an approximation. Thus, your computer requires an algorithm for computing \(\sqrt{51}\) (or any root) to any degree of accuracy. But before implementing such an algorithm, we need to justify that it can indeed approximate \(\sqrt{51}\) to any degree of accuracy. That is, if my algorithm spits out a sequence of numbers \(x_1,x_2,\dots\text{,}\) and we want an approximation that is within \(10^{-10}\) of \(\sqrt{51}\text{,}\) I need to show that for all large enough \(n\text{,}\) \(x_n\) is at most \(10^{-10}\) away from \(\sqrt{51}\text{,}\) and I need this to happen for every degree of accuracy \(\epsilon>0\) that I may require (not just \(10^{-10}\)).

Definition 5.1.1. Sequences and limits.

A sequence of elements of a set \(S\) is a map \(x \colon \NN \to S \text{.}\) For a natural number \(n\text{,}\) we will write \(x_n\) for the value \(x(n)\text{.}\) We will often write the sequence as \((x_n)_{n\geq 1}\) or \((x_n)_{n=1}^{\infty}\) or even just \((x_n)\text{.}\)

Let \((x_{n})_{n\geq 1}\) be a sequence of real numbers, and let \(L\in \RR\text{.}\) We say that \(x_n\) converges to \(L\), and we write “\(\lim_{n\rightarrow\infty} x_{n}=L\)”, or “\(x_{n}\rightarrow L\) as \(n \rightarrow \infty\)”, if for every \(\epsilon>0\text{,}\) there exists \(N \in \NN\) such that for every \(n \gt N\text{,}\) we have

\begin{equation*} |x_{n}-L|\lt \epsilon\text{,} \end{equation*}

or equivalently,

\begin{equation*} L-\epsilon\lt x_n\lt L+\epsilon\text{.} \end{equation*}

We say that a sequence \((x_n)\) is convergent if for some \(L \in \R\text{,}\) it converges to \(L\text{.}\)

One point to be made here is that we aren't so interested in the beginning of our sequences. We're really only interested in what “eventually happens” with the sequence. As a result, we will sometimes find it convenient to consider maps \(x \colon \{n \in \NN : n \geq k\} \to \RR\) for some natural number \(k\) as sequences as well. We may write \((x_n)_{n\geq k}\) or \((x_n)_{n=k}^{\infty}\) for such sequences.

Exercise. Give an example of a convergent sequence. Give an example of a non-convergent sequence.

Show that \(\lim_{n\rightarrow \infty} \frac{1}{n}=0\text{.}\)

We need to show that for all \(\epsilon \gt 0\) there exists a natural number \(N\) so that for every \(n> N\text{,}\) we have \(|\frac{1}{n}-0|\lt \epsilon\text{,}\) which is the same as \(\frac{1}{n}\lt \epsilon\text{.}\) This is equivalent to finding \(N\) so that \(n>N\) implies \(n>\frac{1}{\epsilon}\text{.}\) By the Archimedean property, we may choose a natural number \(N\) such that \(N\epsilon >1\text{.}\) We then see that if \(n> N\text{,}\)

\begin{equation*} \left|\frac{1}{n} -0\right|=\frac{1}{n} \lt \frac{1}{N}= \frac{1}{\frac{1}{\epsilon}}=\epsilon\text{.} \end{equation*}
Note 5.1.3. Note on establishing limits.

Don't expect a nice looking proof to fall into your lap immediately when trying to establish a certain limit. It usually requires some rough work first, which you can then use to help you write your proof.

Show that \(\lim_{n\rightarrow \infty} \frac{5n^2+2n}{n^2+4}=5\text{.}\)

Rough work: We are going to need to show that \(\frac{5n^2+2n}{n^2+4}-5\) is small when \(n\) is large, so it first makes sense to simplify this expression algebraically to see if we can identify when it might be small. Doing this,

\begin{equation*} \frac{5n^2+2n}{n^2+4}-5 =\frac{2n-20}{n^2+4}\text{,} \end{equation*}

and therefore, using the triangle inequality and the fact that \(4\geq0\text{,}\)

\begin{equation*} \left|\frac{5n^2+2n}{n^2+4}-5\right| \leq \left| \frac{2n-20}{n^2+4} \right| \leq \frac{2n}{n^2 + 4} + \frac{20}{n^2 + 4} \leq \frac{2}{n} + \frac{20}{n^2}\text{.} \end{equation*}

This is looking good because all we need to do to make this small is to make both \(2/n\) and \(20/n^2\) small, and this is very reasonable. Now we bring in the \(\epsilon > 0\text{.}\) If \(n\) is such that both \(2/n \lt \epsilon/2\) and \(20/n^2 \lt \epsilon/2\text{,}\) we can deduce that

\begin{equation*} \left|\frac{5n^2+2n}{n^2+4}-5\right| \lt \frac{\epsilon}{2}+ \frac{\epsilon}{2} = \epsilon\text{,} \end{equation*}

which is exactly what we need. What do we need \(n\) to satisfy for this to be true? Well, for \(\frac{2}{n} \lt \frac{\epsilon}{2}\) we need \(n > \frac{4}{\epsilon}\text{,}\) and for \(\frac{20}{n^2} \lt \frac{\epsilon}{2}\) we need \(n > \sqrt{\frac{40}{\epsilon}}\text{.}\) To ensure both of these hold, we need \(n\) to satisfy \(n > \max\left\{\frac{4}{\epsilon}, \sqrt{\frac{40}{\epsilon}}\right\}\text{.}\) So we take \(N = \max\left\{\frac{4}{\epsilon}, \sqrt{\frac{40}{\epsilon}}\right\}\text{.}\)

(All the above was just rough work to help me formulate and write my formal proof. Below is what you want to do when writing your work for marking (and not the rough work). It's fine not to show how you chose your \(N\text{,}\) so long as it works. And there is no single answer for \(N\) — if \(N\) works, then so does \(N'\) for any \(N'>N\text{.}\))

Now for the actual proof:

We claim that \(\lim_{n\rightarrow\infty}\frac{5n^2+2n}{n^2+4}=5\text{.}\) Let \(\epsilon\) be any positive real number. We need to show that there is \(N\in\RR\) so that \(n> N\) implies \(\left|\frac{5n^2+2n}{n^2+4}-5\right|\lt \epsilon\text{.}\) We assert that \(N=\max\left\{\frac{4}{\epsilon}, \sqrt{\frac{40}{\epsilon}}\right\}\) works. Indeed, if \(n> N\text{,}\) then

\begin{equation*} \left|\frac{5n^2+2n}{n^2+4}-5\right| \leq \left| \frac{2n-20}{n^2+4} \right| \leq \frac{2n}{n^2 + 4} + \frac{20}{n^2 + 4} \leq \frac{2}{n} + \frac{20}{n^2} \lt \frac{2}{N} + \frac{20}{N^2} \leq \frac{\epsilon}{2}+ \frac{\epsilon}{2} = \epsilon\text{.} \end{equation*}

Since we have shown we can find such an \(N\) no matter what \(\epsilon>0\) we started with, this proves the claim.

Suppose \(x_{n}\geq 0\) and \(x_{n}\rightarrow x>0\text{,}\) then \(\sqrt{x_{n}}\rightarrow \sqrt{x}\text{.}\)

Discussion: This is a little different from the previous example because now \(x_n\) isn't given in a concrete form, and all we know about it is that \(x_n \rightarrow x\text{,}\) or, more informally, \(|x_n - x|\) is small for all large \(n\text{.}\) Nevertheless the conclusion, \(\sqrt{x_{n}}\rightarrow \sqrt{x}\text{,}\) is similar to what we had before, and the tactic of manipulating the expression \(\sqrt{x_{n}}-\sqrt{x}\) algebraically, and using some inequalities on it until it's apparent why it should also be small, is equally useful. Thus, we want to manipulate \(|\sqrt{x_{n}}-\sqrt{x}|\) so that \(|x_{n}-x|\) appears in the expression, since we hope that if we can make \(|x_{n}-x|\) as small as we want, then we can make \(|\sqrt{x_{n}}-\sqrt{x}|\) small as well.

A standard trick when working with roots is to multiply and divide by the conjugate, and we have

\begin{equation*} |\sqrt{x_{n}}-\sqrt{x}| =\left|(\sqrt{x_{n}}-\sqrt{x})\cdot \frac{\sqrt{x_{n}}+\sqrt{x}}{\sqrt{x_{n}}+\sqrt{x}}\right| =\left|\frac{x_{n}-x}{\sqrt{x_{n}}+\sqrt{x}}\right| =\frac{|x_{n}-x|}{\sqrt{x_{n}}+\sqrt{x}} \leq \frac{|x_{n}-x|}{\sqrt{x}} \end{equation*}

where in the last step we used the facts that \(\sqrt{x_n} \geq 0\) and \(x > 0\text{.}\) This is looking good because we should be able to make \(\frac{|x_{n}-x|}{\sqrt{x}}\) small by taking \(n\) large enough. Indeed, now let's bring in \(\epsilon > 0\text{.}\) Since \(x_n \rightarrow x\text{,}\) we know that there is some \(N \in \RR\) such that whenever \(n > N\text{,}\) we have

\begin{equation*} |x_n - x| \lt \epsilon\sqrt{x}\text{.} \end{equation*}

(This maybe requires a pause for thought. The definition of \(x_n \rightarrow x\) talks about “\(|x_n - x| \lt \epsilon\)” for all large \(n\text{,}\) it doesn't say anything about “\(|x_n - x| \lt \epsilon\sqrt{x}\text{.}\)” But the definition of \(x_n \rightarrow x\) applies to every positive \(\epsilon\) — and thus it applies just equally well to \(\epsilon\sqrt{x}\) in place of \(\epsilon\text{.}\) Remember that \(x\) is fixed in this discussion. The \(N\) coming out depends on \(x\) as well as \(\epsilon\text{,}\) but this is not a problem, as \(x\) is fixed.) So if \(n > N\text{,}\) we have

\begin{equation*} |x_n - x| \leq \frac{|x_n - x|}{\sqrt{x}} \lt \frac{\epsilon\sqrt{x}}{\sqrt{x}} = \epsilon \end{equation*}