## 58. Definition of a limit.

After the discussion which precedes the reader should be in a position to appreciate the general notion of a limit. Roughly we may say that $$\phi(n)$$ tends to a limit $$l$$ as $$n$$ tends to $$\infty$$ if $$\phi(n)$$ is nearly equal to $$l$$ when $$n$$ is large. But although the meaning of this statement should be clear enough after the preceding explanations, it is not, as it stands, precise enough to serve as a strict mathematical definition. It is, in fact, equivalent to a whole class of statements of the type ‘for sufficiently large values of $$n$$, $$\phi(n)$$ differs from $$l$$ by less than $$\epsilon$$’. This statement has to be true for $$\epsilon = .01$$ or $$.0001$$ or any positive number; and for any such value of $$\epsilon$$ it has to be true for any value of $$n$$ after a certain definite value $$n_{0}(\epsilon)$$, though the smaller $$\epsilon$$ is the larger, as a rule, will be this value $$n_{0}(\epsilon)$$.

We accordingly frame the following formal definition:

Definition I. The function $$\phi(n)$$ is said to tend to the limit $$l$$ as $$n$$ tends to $$\infty$$, if, however small be the positive number $$\epsilon$$, $$\phi(n)$$ differs from $$l$$ by less than $$\epsilon$$ for sufficiently large values of $$n$$; that is to say if, however small be the positive number $$\epsilon$$, we can determine a number $$n_{0}(\epsilon)$$ corresponding to $$\epsilon$$, such that $$\phi(n)$$ differs from $$l$$ by less than $$\epsilon$$ for all values of $$n$$ greater than or equal to $$n_{0}(\epsilon)$$.

It is usual to denote the difference between $$\phi(n)$$ and $$l$$, taken positively, by $$|\phi(n) – l|$$. It is equal to $$\phi(n) – l$$ or to $$l – \phi(n)$$, whichever is positive, and agrees with the definition of the modulus of $$\phi(n) – l$$, as given in Ch. III, though at present we are only considering real values, positive or negative.

With this notation the definition may be stated more shortly as follows: ‘if, given any positive number, $$\epsilon$$, however small, we can find $$n_{0}(\epsilon)$$ so that $$|\phi(n) – l| < \epsilon$$ when $$n \geq n_{0}(\epsilon)$$, then we say that $$\phi(n)$$ tends to the limit $$l$$ as $$n$$ tends to $$\infty$$, and write $\lim_{n \to \infty} \phi(n) = l\text{‘.}$

Sometimes we may omit the ‘$$n \to \infty$$’; and sometimes it is convenient, for brevity, to write $$\phi(n) \to l$$.

The reader will find it instructive to work out, in a few simple cases, the explicit expression of $$n_{0}$$ as a function of $$\epsilon$$. Thus if $$\phi({n}) = 1/n$$ then $$l = 0$$, and the condition reduces to $$1/n < \epsilon$$ for $$n \geq n_{0}$$, which is satisfied if $$n_{0} = 1 + [1/\epsilon]$$.1 There is one and only one case in which the same $$n_{0}$$ will do for all values of $$\epsilon$$. If, from a certain value $$N$$ of $$n$$ onwards, $$\phi(n)$$ is constant, say equal to $$C$$, then it is evident that $$\phi(n) – C = 0$$ for $$n \geq N$$, so that the inequality $$|\phi(n) – C| < \epsilon$$ is satisfied for $$n \geq N$$ and all positive values of $$\epsilon$$. And if $$|\phi(n) – l| < \epsilon$$ for $$n \geq N$$ and all positive values of $$\epsilon$$, then it is evident that $$\phi(n) = l$$ when $$n \geq N$$, so that $$\phi(n)$$ is constant for all such values of $$n$$.

## 59.

The definition of a limit may be illustrated geometrically as follows. The graph of $$\phi(n)$$ consists of a number of points corresponding to the values $$n = 1$$, $$2$$, $$3$$, ….

Draw the line $$y = l$$, and the parallel lines $$y = l – \epsilon$$, $$y = l + \epsilon$$ at distance $$\epsilon$$ from it. Then $\lim_{n \to \infty} \phi(n) = l,$ if, when once these lines have been drawn, no matter how close they may be together, we can always draw a line $$x = n_{0}$$, as in the figure, in such a way that the point of the graph on this line, and all points to the right of it, lie between them. We shall find this geometrical way of looking at our definition particularly useful when we come to deal with functions defined for all values of a real variable and not merely for positive integral values.

## 60.

So much for functions of $$n$$ which tend to a limit as $$n$$ tends to $$\infty$$. We must now frame corresponding definitions for functions which, like the functions $$n^{2}$$ or $$-n^{2}$$, tend to positive or negative infinity. The reader should by now find no difficulty in appreciating the point of

Definition II. The function $$\phi(n)$$ is said to tend to $$+\infty$$ (positive infinity) with $$n$$, if, when any number $$\Delta$$, however large, is assigned, we can determine $$n_{0}(\Delta)$$ so that $$\phi(n) > \Delta$$ when $$n \geq n_{0}(\Delta)$$; that is to say if, however large $$\Delta$$ may be, $$\phi(n) > \Delta$$ for sufficiently large values of $$n$$.

Another, less precise, form of statement is ‘if we can make $$\phi(n)$$ as large as we please by sufficiently increasing $$n$$’. This is open to the objection that it obscures a fundamental point, viz. that $$\phi(n)$$ must be greater than $$\Delta$$ for all values of $$n$$ such that $$n \geq n_{0}(\Delta)$$, and not merely for some such values. But there is no harm in using this form of expression if we are clear what it means.

When $$\phi(n)$$ tends to $$+\infty$$ we write $\phi(n) \to +\infty.$ We may leave it to the reader to frame the corresponding definition for functions which tend to negative infinity.

## 61. Some points concerning the definitions.

The reader should be careful to observe the following points.

(1) We may obviously alter the values of $$\phi(n)$$ for any finite number of values of $$n$$, in any way we please, without in the least affecting the behaviour of $$\phi(n)$$ as $$n$$ tends to $$\infty$$. For example $$1/n$$ tends to $$0$$ as $$n$$ tends to $$\infty$$. We may deduce any number of new functions from $$1/n$$ by altering a finite number of its values. For instance we may consider the function $$\phi(n)$$ which is equal to $$3$$ for $$n = 1$$, $$2$$, $$7$$, $$11$$, $$101$$, $$107$$, $$109$$, $$237$$ and equal to $$1/n$$ for all other values of $$n$$. For this function, just as for the original function $$1/n$$, $$\lim\phi(n) = 0$$. Similarly, for the function $$\phi(n)$$ which is equal to $$3$$ if $$n = 1$$, $$2$$, $$7$$, $$11$$, $$101$$, $$107$$, $$109$$, $$237$$, and to $$n^{2}$$ otherwise, it is true that $$\phi(n) \to +\infty$$.

(2) On the other hand we cannot as a rule alter an infinite number of the values of $$\phi(n)$$ without affecting fundamentally its behaviour as $$n$$ tends to $$\infty$$. If for example we altered the function $$1/n$$ by changing its value to $$1$$ whenever $$n$$ is a multiple of $$100$$, it would no longer be true that $$\lim\phi(n) = 0$$. So long as a finite number of values only were affected we could always choose the number $$n_{0}$$ of the definition so as to be greater than the greatest of the values of $$n$$ for which $$\phi(n)$$ was altered. In the examples above, for instance, we could always take $$n_{0} > 237$$, and indeed we should be compelled to do so as soon as our imaginary opponent of § 56 had assigned a value of $$\Delta$$ as small as $$3$$ (in the first example) or a value of $$\Delta$$ as great as $$3$$ (in the second). But now however large $$n_{0}$$ may be there will be greater values of $$n$$ for which $$\phi(n)$$ has been altered.

(3) In applying the test of Definition I it is of course absolutely essential that we should have $$|\phi(n) – l| < \epsilon$$ not merely when $$n = n_{0}$$ but when $$n \geq n_{0}$$,i.e. for $$n_{0}$$ and for all larger values of $$n$$. It is obvious, for example, that, if $$\phi(n)$$ is the function last considered, then given $$\epsilon$$ we can choose $$n_{0}$$ so that $$|\phi(n)| < \epsilon$$ when $$n = n_{0}$$: we have only to choose a sufficiently large value of $$n$$ which is not a multiple of $$100$$. But, when $$n_{0}$$ is thus chosen, it is not true that $$|\phi(n)| < \epsilon$$ when $$n \geq n_{0}$$: all the multiples of $$100$$ which are greater than $$n_{0}$$ are exceptions to this statement.

(4) If $$\phi(n)$$ is always greater than $$l$$, we can replace $$|\phi(n) – l|$$ by $$\phi(n) – l$$. Thus the test whether $$1/n$$ tends to the limit $$0$$ as $$n$$ tends to $$\infty$$ is simply whether $$1/n < \epsilon$$ when $$n \geq n_{0}$$. If however $$\phi(n) = (-1)^{n}/n$$, then $$l$$ is again $$0$$, but $$\phi(n) – l$$ is sometimes positive and sometimes negative. In such a case we must state the condition in the form $$|\phi(n) – l| < \epsilon$$, for example, in this particular case, in the form $$|\phi(n)| < \epsilon$$.

(5) The limit $$l$$ may itself be one of the actual values of $$\phi(n)$$. Thus if $$\phi(n) = 0$$ for all values of $$n$$, it is obvious that $$\lim\phi(n) = 0$$. Again, if we had, in (2) and (3) above, altered the value of the function, when $$n$$ is a multiple of $$100$$, to $$0$$ instead of to $$1$$, we should have obtained a function $$\phi(n)$$ which is equal to $$0$$ when $$n$$ is a multiple of $$100$$ and to $$1/n$$ otherwise. The limit of this function as $$n$$ tends to $$\infty$$ is still obviously zero. This limit is itself the value of the function for an infinite number of values of $$n$$, viz. all multiples of $$100$$.

On the other hand the limit itself need not $$and in general will not$$ be the value of the function for any value of $$n$$. This is sufficiently obvious in the case of $$\phi(n) = 1/n$$. The limit is zero; but the function is never equal to zero for any value of $$n$$.

The reader cannot impress these facts too strongly on his mind. A limit is not a value of the function: it is something quite distinct from these values, though it is defined by its relations to them and may possibly be equal to some of them. For the functions $\phi(n) = 0,\ 1,$ the limit is equal to all the values of $$\phi(n)$$: for $\phi(n) = 1/n,\quad (-1)^{n}/n,\quad 1 + (1/n),\quad 1 + \{(-1)^{n}/n\}$ it is not equal to any value of $$\phi(n)$$: for $\phi(n) = (\sin\tfrac{1}{2}n\pi)/n,\quad 1 + \{(\sin\tfrac{1}{2}n\pi)/n\}$ (whose limits as $$n$$ tends to $$\infty$$ are easily seen to be $$0$$ and $$1$$, since $$\sin\frac{1}{2}n\pi$$ is never numerically greater than $$1$$) the limit is equal to the value which $$\phi(n)$$ assumes for all even values of $$n$$, but the values assumed for odd values of $$n$$ are all different from the limit and from one another.

(6) A function may be always numerically very large when $$n$$ is very large without tending either to $$+\infty$$ or to $$-\infty$$. A sufficient illustration of this is given by $$\phi(n) = (-1)^{n} n$$. A function can only tend to $$+\infty$$ or to $$-\infty$$ if, after a certain value of $$n$$, it maintains a constant sign.

Example XXIII

Consider the behaviour of the following functions of $${n}$$ as $$n$$ tends to $$\infty$$:

1. $$\phi(n) = n^{k}$$, where $$k$$ is a positive or negative integer or rational fraction. If $$k$$ is positive, then $$n^{k}$$ tends to $$+\infty$$ with $$n$$. If $$k$$ is negative, then $$\lim n^{k} = 0$$. If $$k = 0$$, then $$n^{k} = 1$$ for all values of $$n$$. Hence $$\lim n^{k} = 1$$.

The reader will find it instructive, even in so simple a case as this, to write down a formal proof that the conditions of our definitions are satisfied. Take for instance the case of $$k > 0$$. Let $$\Delta$$ be any assigned number, however large. We wish to choose $$n_{0}$$ so that $$n^{k} > \Delta$$ when $$n \geq n_{0}$$. We have in fact only to take for $$n_{0}$$ any number greater than $$\sqrt[k]{\Delta}$$. If $$k = 4$$, then $$n^{4} > 10,000$$ when $$n \geq 11$$, $$n^{4}> 100,000,000$$ when $$n \geq 101$$, and so on.

2. $$\phi(n) = p_{n}$$, where $$p_{n}$$ is the $$n$$th prime number. If there were only a finite number of primes then $$\phi(n)$$ would be defined only for a finite number of values of $$n$$. There are however, as was first shown by Euclid, infinitely many primes. Euclid’s proof is as follows. If there are only a finite number of primes, let them be $$1$$, $$2$$, $$3$$, $$5$$, $$7$$, $$11$$, … $$N$$. Consider the number $$1 + (1 \cdot 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \dots N)$$. This number is evidently not divisible by any of $$2$$, $$3$$, $$5$$, … $$N$$, since the remainder when it is divided by any of these numbers is $$1$$. It is therefore not divisible by any prime save $$1$$, and is therefore itself prime, which is contrary to our hypothesis.

It is moreover obvious that $$\phi(n) > n$$ for all values of $$n$$ (save $$n = 1$$, $$2$$, $$3$$). Hence $$\phi(n) \to +\infty$$.

3. Let $$\phi(n)$$ be the number of primes less than $$n$$. Here again $$\phi(n) \to +\infty$$.

4. $$\phi(n) = [\alpha n]$$, where $$\alpha$$ is any positive number. Here $\phi(n) = 0\quad (0 \leq n < 1 / \alpha),\qquad \phi(n) = 1\quad (1/\alpha \leq n < 2/\alpha),$ and so on; and $$\phi(n) \to +\infty$$.

5. If $$\phi(n) = 1,000,000/n$$, then $$\lim\phi(n) = 0$$: and if $$\psi(n) = n/1,000,000$$, then $$\psi(n) \to +\infty$$. These conclusions are in no way affected by the fact that at first $$\phi(n)$$ is much larger than $$\psi(n)$$, being in fact larger until $$n = 1,000,000$$.

6. $$\phi(n) = 1/\{n – (-1)^{n}\}$$, $$n – (-1)^{n}$$, $$n\{1 – (-1)^{n}\}$$. The first function tends to $$0$$, the second to $$+\infty$$, the third does not tend either to a limit or to $$+\infty$$.

7. $$\phi(n) = (\sin n\theta\pi)/n$$, where $$\theta$$ is any real number. Here $$|\phi(n)| < 1/n$$, since $$|\sin n\theta\pi| \leq 1$$, and $$\lim\phi(n) = 0$$.

8. $$\phi(n) = (\sin n\theta\pi)/\sqrt{n}$$, $$(a\cos^{2} n\theta + b\sin^{2}n\theta)/n$$, where $$a$$ and $$b$$ are any real numbers.

9. $$\phi(n) = \sin n\theta\pi$$. If $$\theta$$ is integral then $$\phi(n) = 0$$ for all values of $$n$$, and therefore $$\lim\phi(n) = 0$$.

Next let $$\theta$$ be rational, e.g. $$\theta = p/q$$, where $$p$$ and $$q$$ are positive integers. Let $$n = aq + b$$ where $$a$$ is the quotient and $$b$$ the remainder when $$n$$ is divided by $$q$$. Then $$\sin(np\pi/q) = (-1)^{ap}\sin(bp\pi/q)$$. Suppose, for example, $$p$$ even; then, as $$n$$ increases from $$0$$ to $$q – 1$$, $$\phi(n)$$ takes the values $0,\quad \sin(p\pi/q),\quad \sin(2p\pi/q),\ \dots\quad \sin\{(q – 1)p\pi/q\}.$ When $$n$$ increases from $$q$$ to $$2q – 1$$ these values are repeated; and so also as $$n$$ goes from $$2q$$ to $$3q – 1$$, $$3q$$ to $$4q – 1$$, and so on. Thus the values of $$\phi(n)$$ form a perpetual cyclic repetition of a finite series of different values. It is evident that when this is the case $$\phi(n)$$ cannot tend to a limit, nor to $$+\infty$$, nor to $$-\infty$$, as $$n$$ tends to infinity.

The case in which $$\theta$$ is irrational is a little more difficult. It is discussed in the next set of examples.

1. Here and henceforward we shall use $$[x]$$ in the sense of Chapt II, as the greatest integer not greater than $$x$$.↩︎