2.4 Convergence: the $\varepsilon$-$N$ definition of a limit

Before giving the formal definition of a limit, let’s try to motivate some of the key ideas by considering another (simpler) example. Recall the graph of the terms of the sequence $(1/n)_{n\in\mathbb{N}}$ from Figure 2.1, which is also reproduced in Figure 2.6. The graph has a horizontal asymptote around the line $y=0$, which reflects the fact that the terms $1/n$ get closer and closer to $0$ as $n$ gets large. Intuitively, we’d like to say the limit of the sequence is $0$, but what do we really mean by this?

Let’s try to be a bit more concrete. After the $100$th term, we will always have $0\leq 1/n<0.01$, so the terms are then within $0.01$ of the limit $0$. After the $1000$th term we will always have $0\leq 1/n<0.001$, so the terms are then within $0.001$ of the limit $0$. In fact, if we prescribe any measure of ‘closeness’ $\varepsilon>0$ (which could be $0.01$, $0.001$, or even $0.00000000000001$), then after the $\lceil 1/\varepsilon\rceil$-term,²² 2 Here $\lceil x\rceil:=\min\{k\in\mathbb{Z}:k\geq x\}$ denotes the ceiling function; in other words, it is the number formed by rounding $x$ up to the nearest integer. we will have $0\leq 1/n<\varepsilon$: see Figure 2.6. Thus,

Consider the sequence $(1/n)_{n\in\mathbb{N}}$ which was informally discussed in Example 2.22. We expect that $1/n\to 0$ as $n\to\infty$, but how do we prove $1/n\to 0$ as $n\to\infty$ working from the $\varepsilon$-$N$ definition?

Walk through. Before presenting the formal proof, we’ll walk through the thought processes involved. For this, we slowly and carefully consider the definition, one quantifier at a time. The first quantifier in Definition 2.23 is for all $\varepsilon>0$. We therefore begin by fixing some arbitrary $\varepsilon>0$ to work with (that is, we prescribe a measure of closeness). A typical way to start a proof is:

Let $\varepsilon>0$ be given.

Now we have fixed $\varepsilon>0$, we move on to the next quantifier. We want to show there exists some $N\in\mathbb{N}$ such that (2.2) holds with $a_{n}:=1/n$ and $a:=0$. This is typically where we need to do the real work, but here most of what we need was already discussed in Example 2.22. Let’s begin by writing out (2.2) explicitly for our choice of sequence and target limit. In particular, $|a_{n}-a|=|1/n-0|=1/n$, and so (2.2) becomes

In other words, we want to find sufficiently large $N\in\mathbb{N}$ such that all terms beyond the $N$th term are tiny: they are smaller than the prescribed $\varepsilon$. How can we do this? Well, if we take $N:=1/\varepsilon$, then whenever $n>N$ we have $1/n<1/N=\varepsilon$, so this choice would work. One small issue is that Definition 2.23 asks that $N$ should be a natural number, and there is no guarantee that the choice $1/\varepsilon$ has this property.³³ 3 The condition that $N$ is a natural number is not an essential part of the definition but more a matter of convention. However, it is a commonly used convention, so we shall stick with it. However, this is easily remedied by taking $N:=\lceil 1/\varepsilon\rceil$, as we did in Example 2.22.

Now we’ve worked out what our choice of $N$ should be, we can write the next line of the proof:

Choose $N:=\lceil 1/\varepsilon\rceil\in\mathbb{N}$.

The next step is to verify that this choice of $N$ works, in the sense that the desired condition (2.2) holds for this choice. However, we chose $N$ precisely with this property in mind, so we can move swiftly on to the next line of the proof:

If $n\in\mathbb{N}$ with $n>N$, then $|1/n-0|=1/n<1/N\leq\varepsilon$.

Finally, it’s helpful to conclude our proof with a wrap up of what we’ve shown:

Thus, by the $\varepsilon$-$N$ definition of a limit, $1/n\to 0$ as $n\to\infty$.

This concludes our walk through of this example. Now we can put the pieces together to come up with our formal proof.

Proof.

Let $\varepsilon>0$ be given and choose $N:=\lceil 1/\varepsilon\rceil\in\mathbb{N}$. If $n\in\mathbb{N}$ with $n>N$, then

$$|1/n-0|=1/n<1/N\leq\varepsilon.$$

Thus, by the $\varepsilon$-$N$ definition of a limit, $1/n\to 0$ as $n\to\infty$. ∎

We claim that $\displaystyle\frac{5n^{2}}{n^{2}+3n}\to 5$ as $n\to\infty$.

For large values of $n$, the term $3n$ on the denominator is much, much smaller than the term $n^{2}$. Therefore, intuitively, $n^{2}+3n$ is relatively close to $n^{2}$ for large values of $n$ and, since $\frac{5n^{2}}{n^{2}}=5$, we expect $\frac{5n^{2}}{n^{2}+3n}\to 5$ as $n\to\infty$. We shall make this intuition precise using the $\varepsilon$-$N$ definition.

We claim that $\sqrt{n+1}-\sqrt{n}\to 0$ as $n\to\infty$.

Intuitively, we expect $\sqrt{n+1}-\sqrt{n}\to 0$ because the graph of the square root function gets ‘flatter and flatter’ as $n$ gets large: see Figure 2.8. Thus, for large value of $n$ there is very little difference between $\sqrt{n+1}$ and $\sqrt{n}$. We shall make this intuition precise using the $\varepsilon$-$N$ definition.

In the proof, similarly to Exercise 2.31, we use the factorisation $x^{2}-y^{2}=(x-y)(x+y)$ for difference of squares to rewrite

It seems like this should give a more complicated expression, but in this case it becomes simpler because for $x=\sqrt{n+1}$ and $y=\sqrt{n}$ we have $x^{2}-y^{2}=1$. This is a bit of a trick, so don’t worry if you didn’t spot it – at least now you know!