## 63. Some general theorems with regard to limits. A. The behaviour of the sum of two functions whose behaviour is known.

Theorem I. If $$\phi(n)$$ and $$\psi(n)$$ tend to limits $$a$$, $$b$$, then $$\phi(n) + \psi(n)$$ tends to the limit $$a + b$$.

This is almost obvious.1 The argument which the reader will at once form in his mind is roughly this: ‘when $$n$$ is large, $$\phi(n)$$ is nearly equal to $$a$$ and $$\psi(n)$$ to $$b$$, and therefore their sum is nearly equal to $$a + b$$’. It is well to state the argument quite formally, however.

Let $$\epsilon$$ be any assigned positive number (e.g. $$.001$$, $$.000 000 1$$, …). We require to show that a number $$n_{0}$$ can be found such that $\begin{equation*} |\phi(n) + \psi(n) – a – b| < \epsilon, \tag{1} \end{equation*}$ when $$n \geq n_{0}$$. Now by a proposition proved in Ch. III (more generally indeed than we need here) the modulus of the sum of two numbers is less than or equal to the sum of their moduli. Thus $|\phi(n) + \psi(n) – a – b| \leq |\phi(n) – a| + |\psi(n) – b|.$ It follows that the desired condition will certainly be satisfied if $$n_{0}$$ can be so chosen that $\begin{equation*} |\phi(n) – a| + |\psi(n) – b| < \epsilon, \tag{2} \end{equation*}$ when $$n \geq n_{0}$$. But this is certainly the case. For since $$\lim\phi(n) = a$$ we can, by the definition of a limit, find $$n_{1}$$ so that $$|\phi(n) – a| < \epsilon’$$ when $$n \geq n_{1}$$, and this however small $$\epsilon’$$ may be. Nothing prevents our taking $$\epsilon’ = \frac{1}{2}\epsilon$$, so that $$|\phi(n) – a| < \frac{1}{2}\epsilon$$ when $$n \geq n_{1}$$. Similarly we can find $$n_{2}$$ so that $$|\psi(n) – b| < \frac{1}{2}\epsilon$$ when $$n \geq n_{2}$$. Now take $$n_{0}$$ to be the greater of the two numbers $$n_{1}$$, $$n_{2}$$. Then $$|\phi(n) – a| < \frac{1}{2}\epsilon$$ and $$|\psi(n) – b| < \frac{1}{2}\epsilon$$ when $$n \geq n_{0}$$, and therefore (2) is satisfied and the theorem is proved.

The argument may be concisely stated thus: since $$\lim\phi(n) = a$$ and $$\lim\psi(n) = b$$, we can choose $$n_{1}$$, $$n_{2}$$ so that $|\phi(n) – a| < \tfrac{1}{2}\epsilon\quad (n \geq n_{1}),\qquad |\psi(n) – b| < \tfrac{1}{2}\epsilon\quad (n \geq n_{2});$ and then, if $$n$$ is not less than either $$n_{1}$$ or $$n_{2}$$, $|\phi(n) + \psi(n) – a – b| \leq |\phi(n) – a| + |{\psi}(n) – b| < \epsilon;$ and therefore $\lim\{\phi(n) + \psi(n)\} = a + b.$

## 64. Results subsidiary to Theorem I.

The reader should have no difficulty in verifying the following subsidiary results.

1. If $$\phi(n)$$ tends to a limit, but $$\psi(n)$$ tends to $$+\infty$$ or to $$-\infty$$ or oscillates finitely or infinitely, then $$\phi(n) + \psi(n)$$ behaves like $$\psi(n)$$.

2. If $$\phi(n) \to +\infty$$, and $$\psi(n) \to +\infty$$ or oscillates finitely, then $$\phi(n) + \psi(n) \to +\infty$$.

In this statement we may obviously change $$+\infty$$ into $$-\infty$$ throughout.

3. If $$\phi(n) \to {+\infty}$$ and $$\psi(n) \to -\infty$$, then $$\phi(n) + \psi(n)$$ may tend either to a limit or to $$+\infty$$ or to $$-\infty$$ or may oscillate either finitely or infinitely.

These five possibilities are illustrated in order by (i) $$\phi(n) = n$$, $$\psi(n) = -n$$, (ii) $$\phi(n) = n^{2}$$, $$\psi(n) = -n$$, (iii) $$\phi(n) = n$$, $$\psi(n) = -n^{2}$$, (iv) $$\phi(n) = n + (-1)^{n}$$, $$\psi(n) = -n$$, (v) $$\phi(n) = n^{2} + (-1)^{n}n$$, $$\psi(n) = -n^{2}$$. The reader should construct additional examples of each case.

4. If $$\phi(n) \to +\infty$$ and $$\psi(n)$$ oscillates infinitely, then $$\phi(n) + \psi(n)$$ may tend to $$+\infty$$ or oscillate infinitely, but cannot tend to a limit, or to $$-\infty$$, or oscillate finitely.

For $$\psi(n) = \{\phi(n) + \psi(n)\} – \phi(n)$$; and, if $$\phi(n) + \psi(n)$$ behaved in any of the three last ways, it would follow, from the previous results, that $$\psi(n) \to -\infty$$, which is not the case. As examples of the two cases which are possible, consider (i) $$\phi(n) = n^{2}$$, $$\psi(n) = (-1)^{n}n$$, (ii) $$\phi(n) = n$$, $$\psi(n) = (-1)^{n}n^{2}$$. Here again the signs of $$+\infty$$ and $$-\infty$$ may be permuted throughout.

5. If $$\phi(n)$$ and $$\psi(n)$$ both oscillate finitely, then $$\phi(n) + \psi(n)$$ must tend to a limit or oscillate finitely.

As examples take $(i) \phi(n) = (-1)^{n},\quad \psi(n) = (-1)^{n+1},\qquad (ii) \phi(n) = \psi(n) = (-1)^{n}.$

6. If $$\phi(n)$$ oscillates finitely, and $$\psi(n)$$ infinitely, then $$\phi(n) + \psi(n)$$ oscillates infinitely.

For $$\phi(n)$$ is in absolute value always less than a certain constant, say $$K$$. On the other hand $$\psi(n)$$, since it oscillates infinitely, must assume values numerically greater than any assignable number (e.g. $$10K$$, $$100K$$, …). Hence $$\phi(n) + \psi(n)$$ must assume values numerically greater than any assignable number ( $$9K$$, $$99K$$, …). Hence $$\phi(n) + \psi(n)$$ must either tend to $$+\infty$$ or $$-\infty$$ or oscillate infinitely. But if it tended to $$+\infty$$ then $\psi(n) = \{\phi(n) + \psi(n)\} – \phi(n)$ would also tend to $$+\infty$$, in virtue of the preceding results. Thus $$\phi(n) + \psi(n)$$ cannot tend to $$+\infty$$, nor, for similar reasons, to $$-\infty$$: hence it oscillates infinitely.

7. If both $$\phi(n)$$ and $$\psi(n)$$ oscillate infinitely, then $$\phi(n) + \psi(n)$$ may tend to a limit, or to $$+\infty$$, or to $$-\infty$$, or oscillate either finitely or infinitely.

Suppose, for instance, that $$\phi(n) = (-1)^{n}n$$, while $$\psi(n)$$ is in turn each of the functions $$(-1)^{n+1}n$$, $$\{1 + (-1)^{n+1}\}n$$, $$-\{1 + (-1)^{n}\}n$$, $$(-1)^{n+1}(n + 1)$$, $$(-1)^{n}n$$. We thus obtain examples of all five possibilities.

The results 1–7 cover all the cases which are really distinct. Before passing on to consider the product of two functions, we may point out that the result of Theorem I may be immediately extended to the sum of three or more functions which tend to limits as $$n\to\infty$$.

## 65. B. The behaviour of the product of two functions whose behaviour is known.

We can now prove a similar set of theorems concerning the product of two functions. The principal result is the following.

Theorem II. If $$\lim\phi(n) = a$$ and $$\lim\psi(n) = b$$, then $\lim\phi(n)\psi(n) = ab.$

Let $\phi(n) = a + \phi_{1}(n),\quad \psi(n) = b + \psi_{1}(n),$ so that $$\lim\phi_{1}(n) = 0$$ and $$\lim\psi_{1}(n) = 0$$. Then $\phi(n)\psi(n) = ab + a\psi_{1}(n) + b\phi_{1}(n) + \phi_{1}(n)\psi_{1}(n).$ Hence the numerical value of the difference $$\phi(n)\psi(n) – ab$$ is certainly not greater than the sum of the numerical values of $$a\psi_{1}(n)$$, $$b\phi_{1}(n)$$, $$\phi_{1}(n)\psi_{1}(n)$$. From this it follows that $\lim\{\phi(n)\psi(n) – ab\} = 0,$ which proves the theorem.

The following is a strictly formal proof. We have $|\phi(n)\psi(n) – ab| \leq |a\psi_{1}(n)| + |b\phi_{1}(n)| + |\phi_{1}(n)| |\psi_{1}(n)|.$ Assuming that neither $$a$$ nor $$b$$ is zero, we may choose $$n_{0}$$ so that $|\phi_{1}(n)| < \tfrac{1}{3}\epsilon/|b|,\quad |\psi_{1}(n)| < \tfrac{1}{3}\epsilon/|a|,$ when $$n \geq n_{0}$$. Then $|\phi(n)\psi(n) – ab| < \tfrac{1}{3}\epsilon + \tfrac{1}{3}\epsilon + \{\tfrac{1}{9}\epsilon^{2}/(|a||b|)\},$ which is certainly less than $$\epsilon$$ if $$\epsilon < \frac{1}{3}|a||b|$$. That is to say we can choose $$n_{0}$$ so that $$|\phi(n)\psi(n) – ab| < \epsilon$$ when $$n \geq n_{0}$$, and so the theorem follows. The reader should supply a proof for the case in which at least one of $$a$$ and $$b$$ is zero.

We need hardly point out that this theorem, like Theorem I, may be immediately extended to the product of any number of functions of $$n$$. There is also a series of subsidiary theorems concerning products analogous to those stated in § 64 for sums. We must distinguish now six different ways in which $$\phi(n)$$ may behave as $$n$$ tends to $$\infty$$. It may (1) tend to a limit other than zero, (2) tend to zero, (3a) tend to $$+\infty$$, (3b) tend to $$-\infty$$, (4) oscillate finitely, (5) oscillate infinitely. It is not necessary, as a rule, to take account separately of (3a) and (3b), as the results for one case may be deduced from those for the other by a change of sign.

To state these subsidiary theorems at length would occupy more space than we can afford. We select the two which follow as examples, leaving the verification of them to the reader. He will find it an instructive exercise to formulate some of the remaining theorems himself.

(i) If $$\phi(n) \to +\infty$$ and $$\psi(n)$$ oscillates finitely, then $$\phi(n)\psi(n)$$ must tend to $$+\infty$$ or to $$-\infty$$ or oscillate infinitely.

Examples of these three possibilities may be obtained by taking $$\phi(n)$$ to be $$n$$ and $$\psi(n)$$ to be one of the three functions $$2 + (-1)^{n}$$, $$-2 – (-1)^{n}$$, $$(-1)^{n}$$.

(ii) If $$\phi(n)$$ and $$\psi(n)$$ oscillate finitely, then $$\phi(n)\psi(n)$$ must tend to a limit $$which may be zero$$ or oscillate finitely.

For examples, take (a) $$\phi(n) = \psi(n) = (-1)^{n}$$, (b) $$\phi(n) = 1 + (-1)^{n}$$, $$\psi(n) = 1 – (-1)^{n}$$, and (c) $$\phi(n) = \cos\frac{1}{3}n\pi$$, $$\psi(n) = \sin\tfrac{1}{3} n\pi$$.

A particular case of Theorem II which is important is that in which $$\psi(n)$$ is constant. The theorem then asserts simply that $$\lim k\phi(n) = ka$$ if $$\lim\phi(n) = a$$. To this we may join the subsidiary theorem that if $$\phi(n) \to +\infty$$ then $$k\phi(n) \to +\infty$$ or $$k\phi(n) \to -\infty$$, according as $$k$$ is positive or negative, unless $$k = 0$$, when of course $$k\phi(n) = 0$$ for all values of $$n$$ and $$\lim k\phi(n) = 0$$. And if $$\phi(n)$$ oscillates finitely or infinitely, then so does $$k\phi(n)$$, unless $$k = 0$$.

## 66. C. The behaviour of the difference or quotient of two functions whose behaviour is known.

There is, of course, a similar set of theorems for the difference of two given functions, which are obvious corollaries from what precedes. In order to deal with the quotient $\frac{\phi(n)}{\psi(n)},$ we begin with the following theorem.

Theorem III. If $$\lim\phi(n) = a$$, and $$a$$ is not zero, then $\lim\frac{1}{\phi(n)} = \frac{1}{a}.$

Let $\phi(n) = a + \phi_{1}(n),$ so that $$\lim\phi_{1}(n) = 0$$. Then $\left|\frac{1}{\phi(n)} – \frac{1}{a}\right| = \frac{|\phi_{1}(n)|}{|a| |a + \phi_{1}(n)|},$ and it is plain, since $$\lim\phi_{1}(n) = 0$$, that we can choose $$n_{0}$$ so that this is smaller than any assigned number $$\epsilon$$ when $$n \geq n_{0}$$.

From Theorems II and III we can at once deduce the principal theorem for quotients, viz.

Theorem IV. If $$\lim\phi(n) = a$$ and $$\lim\psi(n) = b$$, and $$b$$ is not zero, then $\lim\frac{\phi(n)}{\psi(n)} = \frac{a}{b}.$

The reader will again find it instructive to formulate, prove, and illustrate by examples some of the ‘subsidiary theorems’ corresponding to Theorems III and IV.

## 67.

Theorem V. If $$R\{\phi(n), \psi(n), \chi(n), \dots\}$$ is any rational function of $$\phi(n)$$, $$\psi(n)$$, $$\chi(n)$$, …, i.e. any function of the form $P\{\phi(n), \psi(n), \chi(n), \dots\}/Q\{\phi(n), \psi(n), \chi(n), \dots\},$ where $$P$$ and $$Q$$ denote polynomials in $$\phi(n)$$, $$\psi(n)$$, $$\chi(n)$$, …: and if $\lim\phi(n) = a,\quad \lim\psi(n) = b,\quad \lim\chi(n) = c,\ \dots,$ and $Q(a, b, c, \dots) \neq 0;$ then $\lim R\{\phi(n), \psi(n), \chi(n), \dots\} = R(a, b, c, \dots).$

For $$P$$ is a sum of a finite number of terms of the type $A\{\phi(n)\}^{p} \{\psi(n)\}^{q} \dots,$ where $$A$$ is a constant and $$p$$, $$q$$, … positive integers. This term, by Theorem II (or rather by its obvious extension to the product of any number of functions) tends to the limit $$Aa^{p}b^{q}\dots$$, and so $$P$$ tends to the limit $$P(a, b, c, \dots)$$, by the similar extension of Theorem I. Similarly $$Q$$ tends to $$Q(a, b, c, \dots)$$; and the result then follows from Theorem IV.

## 68.

The preceding general theorem may be applied to the following very important particular problem: what is the behaviour of the most general rational function of $$n$$, viz. $S(n) = \frac{a_{0}n^{p} + a_{1}n^{p-1} + \dots + a_{p}} {b_{0}n^{q} + b_{1}n^{q-1} + \dots + b_{q}},$ as $$n$$ tends to $$\infty$$?2

In order to apply the theorem we transform $$S(n)$$ by writing it in the form $n^{p-q}\left\{ \biggl(a_{0} + \frac{a_{1}}{n} + \dots + \frac{a_{p}}{n^{p}}\biggr)\bigg/ \biggl(b_{0} + \frac{b_{1}}{n} + \dots + \frac{b_{q}}{n^{q}}\biggr) \right\}.$ The function in curly brackets is of the form $$R\{\phi(n)\}$$, where $$\phi(n) = 1/n$$, and therefore tends, as $$n$$ tends to $$\infty$$, to the limit $$R(0) = a_{0}/b_{0}$$. Now $$n^{p-q} \to 0$$ if $$p < q$$; $$n^{p-q} = 1$$ and $$n^{p-q} \to 1$$ if $$p = q$$; and $$n^{p-q} \to +\infty$$ if $$p > q$$. Hence, by Theorem II, $\begin{gathered} \lim S(n) = 0\quad (p < q), \\ \lim S(n) = a_{0}/b_{0}\quad (p = q), \\ S(n) \to +\infty\quad (p > q,\ \text{a_{0}/b_{0}positive}), \\ S(n) \to -\infty\quad (p > q,\ \text{a_{0}/b_{0} negative}).\end{gathered}$

Example XXVI

1. What is the behaviour of the functions $\left(\frac{n – 1}{n + 1}\right)^{2},\quad (-1)^{n} \left(\frac{n – 1}{n + 1}\right)^{2},\quad \frac{n^{2} + 1}{n},\quad (-1)^{n} \frac{n^{2} + 1}{n},$ as $$n\to\infty$$?

2. Which (if any) of the functions $\begin{gathered} 1/(\cos^{2}\tfrac{1}{2}n\pi + n\sin^{2}\tfrac{1}{2}n\pi),\quad 1/\{n(\cos^{2}\tfrac{1}{2}n\pi + n\sin^{2}\tfrac{1}{2}n\pi)\}, \\ (n\cos^{2}\tfrac{1}{2}n\pi + \sin^{2}\tfrac{1}{2}n\pi)/ \{n(\cos^{2}\tfrac{1}{2}n\pi + n\sin^{2}\tfrac{1}{2}n\pi)\}\end{gathered}$ tend to a limit as $$n \to \infty$$?

3. Denoting by $$S(n)$$ the general rational function of $$n$$ considered above, show that in all cases $\lim\frac{S(n + 1)}{S(n)} = 1,\quad \lim\frac{S\{n + (1/n)\}}{S(n)} = 1.$

1. There is a certain ambiguity in this phrase which the reader will do well to notice. When one says ‘such and such a theorem is almost obvious’ one may mean one or other of two things. One may mean ‘it is difficult to doubt the truth of the theorem’, ‘the theorem is such as common-sense instinctively accepts’, as it accepts, for example, the truth of the propositions ‘$$2 + 2 = 4$$’ or ‘the base-angles of an isosceles triangle are equal’. That a theorem is ‘obvious’ in this sense does not prove that it is true, since the most confident of the intuitive judgments of common sense are often found to be mistaken; and even if the theorem is true, the fact that it is also ‘obvious’ is no reason for not proving it, if a proof can be found. The object of mathematics is to prove that certain premises imply certain conclusions; and the fact that the conclusions may be as ‘obvious’ as the premises never detracts from the necessity, and often not even from the interest of the proof.But sometimes (as for example here) we mean by ‘this is almost obvious’ something quite different from this. We mean ‘a moment’s reflection should not only convince the reader of the truth of what is stated, but should also suggest to him the general lines of a rigorous proof’. And often, when a statement is ‘obvious’ in this sense, one may well omit the proof, not because the proof is in any sense unnecessary, but because it is a waste of time and space to state in detail what the reader can easily supply for himself.↩︎
2. We naturally suppose that neither $$a_{0}$$ nor $$b_{0}$$ is zero.↩︎