## Introduction

First, we explain the underlying basic idea of the epsilon-N definition of the limits of sequences through the parable of an airplane approaching an airport.

Consider a situation in which an airplane departing from New York approaches the destination Tokyo as time passes, as shown Figure 1. Here, for the sake of simplicity, we suppose that the airplane stops moving after arrival at Tokyo airport.

Assume that the airplane arrives at the Tokyo airport at 9:54 on schedule. As the arrival time approaches, the airplane approaches Tokyo airport. As seen in Figure 1, if we set an area within 20 km centered at Tokyo airport, then there exists a time when the airplane enters this area; it is 9:32. Similarly, if we set an area within 10 km centered at Tokyo airport, then there exists a time, 9:43, when the airplane enters this area. Since the airplane approaches Tokyo airport, we should consider a narrower area around the airport. If we set an area within 2 km centered at Tokyo airport, then there exists a time, 9:50, when the airplane enters the area within 2 km. These facts are generally described as follows:

Here, \varepsilon denotes the Greek letter called “epsilon”. Moreover, the time T when the airplane enters the area within \varepsilon km centered at Tokyo airport is determined by \varepsilon. That is, the time T is a function of \varepsilon, denoted by T(\varepsilon). In fact, the cases mentioned above are expressed by

\ \ \ T(20) = 9:32, \ \ \ T(10) = 9:43, \ \ \ T(2) = 9:50.

Furthermore, we can set any small area around Tokyo airport because the airplane approaches the airport. In other words, we can choose \varepsilon arbitrarily.

Is T(\varepsilon) a function? I think that a function has a concrete

expression such as f(x) = x^2 and y = x+1.

Yes, the time T is a function of \varepsilon. In fact, the terminology “function” means a relationship where each input has a single output. It is not necessary to describe a function by a concrete expression.

Is it possible that we cannot define the value of T(\varepsilon)? For example, if the airplane cannot approach Tokyo airport due to a dense fog, how can I consider this situation?

For example, suppose that a dense fog covers an area within 3 km centered at Tokyo airport, and that the airplane cannot enter this area. In this case, we cannot determine the value of T for \varepsilon = 3. As a result, the function T(\varepsilon) does not exist for all \varepsilon. This implies that the airplane cannot approach Tokyo airport.

We are allowed to choose any value of \varepsilon. Could I choose \varepsilon = 100,000 ?

Although you could choose \varepsilon = 100,000, it would be meaningless. In fact, the distance between New York and Tokyo is 10,864 km. Since the airplane approaches Tokyo airport, it is essential to choose as small a value of \varepsilon as possible. In other words, there is no problem in imposing the restriction that \varepsilon should be sufficiently small.

Next, we describe the airplane’s entry into an area within \varepsilon km centered at Tokyo airport. Since the airplane enters this area at a time T(\varepsilon) as shown in Figure 1, we can consider that the airplane is included in the area within \varepsilon km centered at Tokyo airport after the time T(\varepsilon). Therefore, for all time t with t \geq T(\varepsilon), we have

\ \ \ \ | \ \text{the airplane} \ - \ \text{the Tokyo airport} \ | \ < \ \varepsilon.

Here, t \geq T(\varepsilon) means that t is a time after the time T(\varepsilon). Moreover, | \ \text{the airplane} \ - \ \text{the Tokyo airport} \ | denotes the distance between the airplane and Tokyo airport, in the same way that the distance between a and b on the real line is denoted by | \ a \ - \ b \ |. Hence, | \ \text{the airplane} \ - \ \text{the Tokyo airport} \ | \ < \ \varepsilon means that the distance between the airplane and Tokyo airport is smaller than \varepsilon km. Thus, we see that the following statement holds:

Although this type of formal sentence is never used in our daily conversation, it precisely states that the airplane approaches Tokyo airport as time passes. Moreover, it should be noted that T is a function of \varepsilon. The existence of the function T(\varepsilon) guarantees that the airplane approaches Tokyo airport. Furthermore, the function T(\varepsilon) indicates the velocity of the airplane approaching Tokyo airport. For example, if a 20-minite delay occurs due to a disturbance in the air flow around Tokyo airport, then T(\varepsilon) would be given by

\ \ \ T(20) = 9:40, \ \ \ T(10) = 9:56, \ \ \ T(2) = 10:08,

which suggests that the airplane is slowing down.

I think that the time T(\varepsilon) shifts as the distance \varepsilon decreases.

Indeed. It is quite natural because the airplane approaches Tokyo airport as time passes.

I wonder whether the arrival time of the airplane does not appear in the above explanation.

Certainly. We can give an explanation considering the arrival time, which is related to the definition of the continuity of functions. But, in this site, we explain the limits and continuity of functions after the limits of sequences. Therefore, we gave the explanation with no consideration of the arrival time, which is useful to define the limit of a sequence in the next section.

## Limits of Sequences

It is easy to understand the definition of the limit of a sequence if you understand the parable of the airplane explained in the previous section.

An enumerated collection of real numbers such as

1, \ \frac{1}{2}, \ \frac{1}{3}, \ \frac{1}{4}, \ \cdots

is called a sequence. In general, a sequence is denoted by

a_1, \ a_2, \ a_3, \ \cdots, \ a_n, \ \cdots

or by \{ a_n \}, where a_n is called the n-th term. For example, the n-th term of the above sequence is given by

a_n = \frac{1}{n}.

Note that a_n can be regarded as a function of n, where the variable n is called an index.

As for the sequence given by a_n = \displaystyle\frac{1}{n}, the value of a_n decreases and eventually goes to 0 as the index n increases. We may consider that this sequence converges to the limit value 0 on the basis of our naive intuition. However, such intuitive considerations do not always lead to correct results. Here, we propose a rigorous definition of the limit of a sequence in the same way that was done in the parable of the airplane in the previous section.

Consider the following two statements:

(A) An airplane approaches Tokyo airport as time passes.

(B) A sequence infinitely approaches the limit value as the index increases.

Comparing these statements, we find the following correspondences:

airplane \to sequence, Tokyo airport \to limit value, time \to index.

Recalling the precise statement in the previous section that an airplane continuously approaches Tokyo airport as time passes, we propose a rigorous definition of the limit of a sequence as follows:

When \{ a_n \} satisfies this condition, we say that \{ a_n \} converges to \alpha, which is denoted by \displaystyle\lim_{ n \to \infty} a_n = \alpha. Moreover, \alpha is called the limit value of \{ a_n \}. It should be noted that N is a function of \varepsilon. The existence of the function N(\varepsilon) guarantees the convergence of a sequence.

According to my textbook, the definition of the limit of a sequence is given as follows: For any \varepsilon > 0, there exists a natural number N such that for all n \geq N, we have | \ a_n \ - \ \alpha \ | \ < \ \varepsilon. Why is the number N not represented as N(\varepsilon)?

The author of your book would completely understand that N depends on \varepsilon, which implies that N is a function of \varepsilon. I think that the author unconsciously uses this crucial fact. In contrast, beginners using the rigorous definition of the limit of a sequence often overlook that N is a function of \varepsilon. Therefore, you should write N(\varepsilon) while you are unfamilar with the arguments based on this definition.

## Simple Examples

We now show that the sequence

1, \ \frac{1}{2}, \ \frac{1}{3}, \ \frac{1}{4}, \ \cdots

given by a_n = \displaystyle\frac{1}{n} converges to 0. That is, we will show that for any \varepsilon > 0, there exists a natural number N(\varepsilon) such that for all n \geq N(\varepsilon), we have

| \ \frac{1}{n} \ - \ 0 \ | \ < \ \varepsilon.

Note that this inequality is equivalent to

\frac{1}{n} \ < \ \varepsilon; \ \ \text{i.e.,} \ \ n > \frac{1}{\varepsilon}.

Let \varepsilon = 0.01. Noting that 1 \div 0.01 = 100, we set N = N(0.01) = 101. Then, for all n \geq 101, we have

| \ \frac{1}{n} \ - \ 0 \ | \ = \ | \ \frac{1}{n} \ | \ \leq \ | \ \frac{1}{101} \ | \ < \ 0.01.

Similarly, for \varepsilon = 0.003, noting that 1 \div 0.003 = 333.333333 \cdots, we set N = N(0.003) = 334. Then, for all n \geq 334, we have

| \ \frac{1}{n} \ - \ 0 \ | \ = \ | \ \frac{1}{n} \ | \ \leq \ | \ \frac{1}{334} \ | \ < \ 0.003.

In general, for any \varepsilon > 0, we set N(\varepsilon) as the minimum of natural numbers greater than \displaystyle\frac{1}{\varepsilon}. Then, noting that \displaystyle\frac{1}{N(\varepsilon)} < \varepsilon by N(\varepsilon) > \displaystyle\frac{1}{\varepsilon}, for all n \geq N(\varepsilon), we have

| \ \frac{1}{n} \ - \ 0 \ | \ = \ | \ \frac{1}{n} \ | \ \leq \ | \ \frac{1}{N(\varepsilon)} \ | \ < \ \varepsilon.

Thus, we see that \{ a_n \} converges to 0.

We set N(\varepsilon) as the minimum of natural numbers greater than \displaystyle\frac{1}{\varepsilon}. I wonder whether the minimum exists or not. Is it possible to prove the existence of the minimum?

Considering the cases \varepsilon = 0.01 and \varepsilon = 0.003, we intuitively feel that the minimum exists. However, to prove the existence of the minimum, we must admit the following assertions: (1) for any x > 0, there exists a natural number greater than x; (2) there exists the minimum of a non-empty set consisting of natural numbers. There would be no doubt about these assertions. Here, we should not go any further into their rigorous justification.

In the same way as in the previous example, we can show that the sequnece

1, \ \frac{1}{4}, \ \frac{1}{9}, \ \frac{1}{16}, \ \cdots

given by b_n = \displaystyle\frac{1}{n^2} converges to 0. That is, for any \varepsilon > 0, there exists a natural number N(\varepsilon) such that for all n \geq N(\varepsilon), we have

| \ \frac{1}{n^2} \ - \ 0 \ | \ < \ \varepsilon

In fact, noting that

\frac{1}{n^2} \ < \ \varepsilon; \ \ \text{i.e.,} \ \ n > \frac{1}{\sqrt{\varepsilon}},

we recall the argument in the previous example. For any \varepsilon > 0, we set N(\varepsilon) as the minimum of natural numbers greater than \displaystyle\frac{1}{\sqrt{\varepsilon}}. Then, noting that {\large(}N(\varepsilon){\large)}^2 > \displaystyle\frac{1}{\varepsilon}, for all n \geq N(\varepsilon), we have

| \ \frac{1}{n^2} \ - \ 0 \ | \ = \ | \ \frac{1}{n^2} \ | \ \leq \ | \ \frac{1}{{\large(}N(\varepsilon){\large)}^2} \ | \ < \ \varepsilon.

Thus, we see that \{ b_n \} converges to 0.

I think that the value of N(\varepsilon) increases as \varepsilon decreases.

That’s right. As seen in the parable of the airplane, in general, the value of N(\varepsilon) increases as \varepsilon decreases.

Finally, we compare the velocities of convergence to 0 for \{ a_n \} and \{ b_n \}. To do so, we compare the following two functions:

N = N_1(\varepsilon), \ \ \ N_1(\varepsilon) \ \text{being the minimum of natural numbers greater than} \ \frac{1}{\varepsilon}

and

N = N_2(\varepsilon), \ \ \ N_2(\varepsilon) \ \text{being the minimum of natural numbers greater than} \ \frac{1}{\sqrt{\varepsilon}},

which are used in the proofs of the convergence of \{ a_n \} and \{ b_n \}, respectively.

For example, let \varepsilon = 0.01. Noting that N_1(0.01) = 101, we see that for all n \geq 101, | a_n | < 0.01 holds. That is, we must take n satisfying n \geq 101 in order to include a_n in a ball (interval) with radius 0.01 and center 0. In contrast, noting that N_2(0.01) = 11, we see that for all n \geq 11, | b_n | < 0.01 holds. Therefore, it is sufficient to take n satisfying n \geq 11 in order to include b_n in the same ball (interval). Although we consider only a particular case for \varepsilon = 0.01, we can conclude that \{ b_n \} converges to 0 faster than \{ a_n \}. In fact, when 0 < \varepsilon < 1, by \sqrt{\varepsilon} > \varepsilon, we have

\frac{1}{\sqrt{\varepsilon}} < \frac{1}{\varepsilon},

which suggests N_1(\varepsilon) > N_2(\varepsilon).

Is it enough to take \varepsilon satisfying 0 < \varepsilon < 1. Is it unnecessary to consider the case for any \varepsilon > 0?

You have a similar question in the parable of the airplane. When we examine the convergence of sequences, it is crucial to take \varepsilon as small as possible. There is no problem in imposing the restriction \varepsilon < 1.

Exercise 1 Show that the sequence given by c_n = \displaystyle\frac{1}{\sqrt{n}} converges to 0. Moreover, compare the velocities of convergence to 0 for \{ a_n \}, \{ b_n \}, and \{ c_n \}.

Good job! Let’s go to the next chapter.