## 93. Limits as $$x$$ tends to $$0$$.

Let $$\phi(x)$$ be such a function of $$x$$ that $$\lim\limits_{x \to \infty} \phi(x) = l$$, and let $$y = 1/x$$. Then $\phi(x) = \phi(1/y) = \psi(y),$ say. As $$x$$ tends to $$\infty$$, $$y$$ tends to the limit $$0$$, and $$\psi(y)$$ tends to the limit $$l$$.

Let us now dismiss $$x$$ and consider $$\psi(y)$$ simply as a function of $$y$$. We are for the moment concerned only with those values of $$y$$ which correspond to large positive values of $$x$$, that is to say with small positive values of $$y$$. And $$\psi(y)$$ has the property that by making $$y$$ sufficiently small we can make $$\psi(y)$$ differ by as little as we please from $$l$$. To put the matter more precisely, the statement expressed by $$\lim\phi(x) = l$$ means that, when any positive number $$\epsilon$$, however small, is assigned, we can choose $$x_{0}$$ so that $$|\phi(x) – l| < \epsilon$$ for all values of $$x$$ greater than or equal to $$x_{0}$$. But this is the same thing as saying that we can choose $$y_{0} = 1/x_{0}$$ so that $$|\psi(y) – l| < \epsilon$$ for all positive values of $$y$$ less than or equal to $$y_{0}$$.

We are thus led to the following definitions:

A. If, when any positive number $$\epsilon$$, however small, is assigned, we can choose $$y_{0}(\epsilon)$$ so that $|\phi(y) – l| < \epsilon$ when $$0 < y \leq y_{0}(\epsilon)$$, then we say that $$\phi(y)$$ tends to the limit $$l$$ as $$y$$ tends to $$0$$ by positive values, and we write $\lim_{y \to +0} \phi(y) = l.$

B. If, when any number $$\Delta$$, however large, is assigned, we can choose $$y_{0}(\Delta)$$ so that $\phi(y) > \Delta$ when $$0 < y \leq y_{0}(\Delta)$$, then we say that $$\phi(y)$$ tends to $$\infty$$ as $$y$$ tends to $$0$$ by positive values, and we write $\phi(y) \to \infty.$

We define in a similar way the meaning of ‘$$\phi(y)$$ tends to the limit $$l$$ as $$y$$ tends to $$0$$ by negative values’, or ‘$$\lim\phi(y) = l$$ when $$y \to -0$$’. We have in fact only to alter $$0 < y \leq y_{0}(\epsilon)$$ to $$-y_{0}(\epsilon) \leq y < 0$$ in definition A. There is of course a corresponding analogue of definition B, and similar definitions in which $\phi(y) \to -\infty$ as $$y \to +0$$ or $$y \to -0$$.

If $$\lim\limits_{y \to +0} \phi(y) = l$$ and $$\lim\limits_{y \to -0} \phi(y) = l$$, we write simply $\lim_{y \to 0} \phi(y) = l.$ This case is so important that it is worth while to give a formal definition.

If, when any positive number $$\epsilon$$, however small, is assigned, we can choose $$y_{0}(\epsilon)$$ so that, for all values of $$y$$ different from zero but numerically less than or equal to $$y_{0}(\epsilon)$$, $$\phi(y)$$ differs from $$l$$ by less than $$\epsilon$$, then we say that $$\phi(y)$$ tends to the limit $$l$$ as $$y$$ tends to $$0$$, and write $\lim_{y \to 0} \phi(y) = l.$

So also, if $$\phi(y) \to \infty$$ as $$y \to +0$$ and also as $$y \to -0$$, we say that $$\phi(y) \to \infty$$ as $$y \to 0$$. We define in a similar manner the statement that $$\phi(y) \to -\infty$$ as $$y \to 0$$.

Finally, if $$\phi(y)$$ does not tend to a limit, or to $$\infty$$, or to $$-\infty$$, as $$y \to +0$$, we say that $$\phi(y)$$ oscillates as $$y \to +0$$, finitely or infinitely as the case may be; and we define oscillation as $$y \to -0$$ in a similar manner.

The preceding definitions have been stated in terms of a variable denoted by $$y$$: what letter is used is of course immaterial, and we may suppose $$x$$ written instead of $$y$$ throughout them.

## 94. Limits as $$x$$ tends to $$a$$.

Suppose that $$\phi(y) \to l$$ as $$y \to 0$$, and write $y = x – a,\quad \phi(y) = \phi(x – a) = \psi(x).$ If $$y \to 0$$ then $$x \to a$$ and $$\psi(x) \to l$$, and we are naturally led to write $\lim_{x \to a} \psi(x) = l,$ or simply $$\lim\psi(x) = l$$ or $$\psi(x) \to l$$, and to say that $$\psi(x)$$ tends to the limit $$l$$ as $$x$$ tends to $$a$$. The meaning of this equation may be formally and directly defined as follows:

if, given $$\epsilon$$, we can always determine $$\delta(\epsilon)$$ so that $|\phi(x) – l| < \epsilon$ when $$0 < |x – a| \leq \delta(\epsilon)$$, then $\lim_{x \to a} \phi(x) = l.$

By restricting ourselves to values of $$x$$ greater than $$a$$, i.e. by replacing $$0 < |x – a| \leq \delta(\epsilon)$$ by $$a < x \leq a + \delta(\epsilon)$$, we define ‘$$\phi(x)$$ tends to $$l$$ when $$x$$ approaches $$a$$ from the right’, which we may write as $\lim_{x \to a+0} \phi(x) = l.$ In the same way we can define the meaning of $\lim_{x \to a-0} \phi(x) = l.$ Thus $$\lim\limits_{x \to a} \phi(x) = l$$ is equivalent to the two assertions $\lim_{x \to a+0} \phi(x) = l,\quad \lim_{x \to a-0} \phi(x) = l.$

We can give similar definitions referring to the cases in which $$\phi(x) \to \infty$$ or $$\phi(x) \to -\infty$$ as $$x \to a$$ through values greater or less than $$a$$; but it is probably unnecessary to dwell further on these definitions, since they are exactly similar to those stated above in the special case when $$a = 0$$, and we can always discuss the behaviour of $$\phi(x)$$ as $$x \to a$$ by putting $$x – a = y$$ and supposing that $$y \to 0$$.

## 95. Steadily increasing or decreasing functions.

If there is a number $$\delta$$ such that $$\phi(x’) \leq \phi(x”)$$ whenever $$a – \delta < x’ < x” < a + \delta$$, then $$\phi(x)$$ will be said to increase steadily in the neighbourhood of $$x = a$$.

Suppose first that $$x < a$$, and put $$y = 1/(a – x)$$. Then $$y \to \infty$$ as $$x \to a-0$$, and $$\phi(x) = \psi(y)$$ is a steadily increasing function of $$y$$, never greater than $$\phi(a)$$. It follows from § 92 that $$\phi(x)$$ tends to a limit not greater than $$\phi(a)$$. We shall write $\lim_{x \to a+0} \phi(x) = \phi(a+0).$ We can define $$\phi(a-0)$$ in a similar manner; and it is clear that $\phi(a-0) \leq \phi(a) \leq \phi(a+0).$ It is obvious that similar considerations may be applied to decreasing functions.

If $$\phi(x’) < \phi(x”)$$, the possibility of equality being excluded, whenever $$a – \delta < x’ < x” < a + \delta$$, then $$\phi(x)$$ will be said to be steadily increasing in the stricter sense.

## 96. Limits of indetermination and the principle of convergence.

All of the argument of §§ 80-84 may be applied to functions of a continuous variable $$x$$ which tends to a limit $$a$$. In particular, if $$\phi(x)$$ is bounded in an interval including $$a$$ (i.e. if we can find $$\delta$$, $$H$$, and $$K$$ so that $$H < \phi(x) < K$$ when $$a – \delta \leq x \leq a +\delta$$).1 then we can define $$\lambda$$ and $$\Lambda$$, the lower and upper limits of indetermination of $$\phi(x)$$ as $$x \to a$$, and prove that the necessary and sufficient condition that $$\phi(x) \to l$$ as $$x \to a$$ is that $$\lambda = \Lambda = l$$. We can also establish the analogue of the principle of convergence, prove that the necessary and sufficient condition that $$\phi(x)$$ should tend to a limit as $$x \to a$$ is that, when $$\epsilon$$ is given, we can choose $$\delta(\epsilon)$$ so that $$|\phi(x_{2}) – \phi(x_{1})| < \epsilon$$ when $$0 < |x_{2} – a| < |x_{1} – a| \leq \delta(\epsilon)$$.

Example XXXV

1. If $\phi(x) \to l,\quad \psi(x) \to l’,$ as $$x \to a$$, then $$\phi(x) + \psi(x) \to l + l’$$, $$\phi(x)\psi(x) \to ll’$$, and $$\phi(x)/\psi(x) \to l/l’$$, unless in the last case $$l’ = 0$$.

[We saw in § 91 that the theorems of Ch. IV, § 63 et seq. hold also for functions of $$x$$ when $$x \to \infty$$ or $$x \to -\infty$$. By putting $$x = 1/y$$ we may extend them to functions of $$y$$, when $$y \to 0$$, and by putting $$y = z – a$$ to functions of $$z$$, when $$z \to a$$.

The reader should however try to prove them directly from the formal definition given above. Thus, in order to obtain a strict direct proof of the first result he need only take the proof of Theorem I of § 63 and write throughout $$x$$ for $$n$$, $$a$$ for $$\infty$$ and $$0 < |x – a| \leq \delta$$ for $$n \geq n_{0}$$.]

2. If $$m$$ is a positive integer then $$x^{m} \to 0$$ as $$x \to 0$$.

3. If $$m$$ is a negative integer then $$x^{m} \to +\infty$$ as $$x \to +0$$, while $$x^{m} \to -\infty$$ or $$x^{m} \to +\infty$$ as $$x \to -0$$, according as $$m$$ is odd or even. If $$m = 0$$ then $$x^{m} = 1$$ and $$x^{m} \to 1$$.

4. $$\lim\limits_{x \to 0} (a + bx + cx^{2} + \dots + kx^{m}) = a$$.

5. $$\lim\limits_{x \to 0} \left\{(a + bx + \dots + kx^{m})/(\alpha + \beta x + \dots + \kappa x^{\mu})\right\} = a/\alpha$$, unless $$\alpha = 0$$. If $$\alpha = 0$$ and $$a \neq 0$$, $$\beta \neq 0$$, then the function tends to $$+\infty$$ or $$-\infty$$, as $$x \to +0$$, according as $$a$$ and $$\beta$$ have like or unlike signs; the case is reversed if $$x \to -0$$. The case in which both $$a$$ and $$\alpha$$ vanish is considered in Ex. XXXVI 5. Discuss the cases which arise when $$a \neq 0$$ and more than one of the first coefficients in the denominator vanish.

6. $$\lim\limits_{x \to a} x^{m} = a^{m}$$, if $$m$$ is any positive or negative integer, except when $$a = 0$$ and $$m$$ is negative. [If $$m > 0$$, put $$x = y + a$$ and apply Ex. 4. When $$m < 0$$, the result follows from Ex. 1 above. It follows at once that $$\lim P(x) = P(a)$$, if $$P(x)$$ is any polynomial.]

7. $$\lim\limits_{x \to a} R(x) = R(a)$$, if $$R$$ denotes any rational function and $$a$$ is not one of the roots of its denominator.

8. Show that $$\lim\limits_{x \to a} x^{m} = a^{m}$$ for all rational values of $$m$$, except when $$a = 0$$ and $$m$$ is negative. [This follows at once, when $$a$$ is positive, from the inequalities (9) or (10) of § 74. For $$|x^{m} – a^{m}| < H|x – a|$$, where $$H$$ is the greater of the absolute values of $$mx^{m-1}$$ and $$ma^{m-1}$$ (cf. Ex. XXVIII. 4). If $$a$$ is negative we write $$x = -y$$ and $$a = -b$$. Then $\lim x^{m} = \lim (-1)^{m}y^{m} = (-1)^{m}b^{m} = a^{m}.]$

## 97.

The reader will probably fail to see at first that any proof of such results as those of Exs. 4, 5, 6, 7, 8 above is necessary. He may ask ‘why not simply put $$x = 0$$, or $$x = a$$? Of course we then get $$a$$, $$a/\alpha$$, $$a^{m}$$, $$P(a)$$, $$R(a)$$’. It is very important that he should see exactly where he is wrong. We shall therefore consider this point carefully before passing on to any further examples.

The statement $\lim_{x \to 0} \phi(x) = l$ is a statement about the values of $$\phi(x)$$ when $$x$$ has any value distinct from but differing by little from zero.2 It is not a statement about the value of $$\phi(x)$$ when $$x = 0$$. When we make the statement we assert that, when $$x$$ is nearly equal to zero, $$\phi(x)$$ is nearly equal to $$l$$. We assert nothing whatever about what happens when $$x$$ is actually equal to $$0$$. So far as we know, $$\phi(x)$$ may not be defined at all for $$x = 0$$; or it may have some value other than $$l$$. For example, consider the function defined for all values of $$x$$ by the equation $$\phi(x) = 0$$. It is obvious that $\begin{equation*} \lim\phi(x) = 0. \tag{1} \end{equation*}$ Now consider the function $$\psi(x)$$ which differs from $$\phi(x)$$ only in that $$\psi(x) = 1$$ when $$x = 0$$. Then $\begin{equation*} \lim\psi(x) = 0, \tag{2} \end{equation*}$ for, when $$x$$ is nearly equal to zero, $$\psi(x)$$ is not only nearly but exactly equal to zero. But $$\psi(0) = 1$$. The graph of this function consists of the axis of $$x$$, with the point $$x = 0$$ left out, and one isolated point, viz. the point $$(0, 1)$$. The equation expresses the fact that if we move along the graph towards the axis of $$y$$, from either side, then the ordinate of the curve, being always equal to zero, tends to the limit zero. This fact is in no way affected by the position of the isolated point $$(0, 1)$$.

The reader may object to this example on the score of artificiality: but it is easy to write down simple formulae representing functions which behave precisely like this near $$x = 0$$. One is $\psi(x) = [1 – x^{2}],$ where $$[1 – x^{2}]$$ denotes as usual the greatest integer not greater than $$1 – x^{2}$$. For if $$x = 0$$ then $$\psi(x) = [1] = 1$$; while if $$0 < x < 1$$, or $$-1 < x < 0$$, then $$0 < 1 – x^{2} < 1$$ and so $$\psi(x) = [1 – x^{2}] = 0$$.

Or again, let us consider the function $y = x/x$ already discussed in Ch. II, § 24, (2). This function is equal to $$1$$ for all values of $$x$$ save $$x = 0$$. It is not equal to $$1$$ when $$x = 0$$: it is in fact not defined at all for $$x = 0$$. For when we say that $$\phi(x)$$ is defined for $$x = 0$$ we mean (as we explained in Ch. II, l.c.) that we can calculate its value for $$x = 0$$ by putting $$x = 0$$ in the actual expression of $$\phi(x)$$. In this case we cannot. When we put $$x = 0$$ in $$\phi(x)$$ we obtain $$0/0$$, which is a meaningless expression. The reader may object ‘divide numerator and denominator by $$x$$’. But he must admit that when $$x = 0$$ this is impossible. Thus $$y = x/x$$ is a function which differs from $$y = 1$$ solely in that it is not defined for $$x = 0$$. None the less $\lim(x/x) = 1,$ for $$x/x$$ is equal to $$1$$ so long as $$x$$ differs from zero, however small the difference may be.

Similarly $$\phi(x) = \{(x + 1)^{2} – 1\}/x = x + 2$$ so long as $$x$$ is not equal to zero, but is undefined when $$x = 0$$. None the less $$\lim\phi(x) = 2$$.

On the other hand there is of course nothing to prevent the limit of $$\phi(x)$$ as $$x$$ tends to zero from being equal to $$\phi(0)$$, the value of $$\phi(x)$$ for $$x = 0$$. Thus if $$\phi(x) = x$$ then $$\phi(0) = 0$$ and $$\lim\phi(x) = 0$$. This is in fact, from a practical point of view, from the point of view of what most frequently occurs in applications, the ordinary case.

Example XXXVI

1. $$\lim\limits_{x \to a} (x^{2} – a^{2})/(x – a) = 2a$$.

2. $$\lim\limits_{x \to a} (x^{m} – a^{m})/(x – a) = ma^{m-1}$$, if $$m$$ is any integer (zero included).

3. Show that the result of Ex. 2 remains true for all rational values of $$m$$, provided $$a$$ is positive. [This follows at once from the inequalities (9) and (10) of § 74.]

4. $$\lim\limits_{x \to 1} (x^{7} – 2x^{5} + 1)/(x^{3} – 3x^{2} + 2) = 1$$. [Observe that $$x – 1$$ is a factor of both numerator and denominator.]

5. Discuss the behaviour of $\phi(x) = (a_{0}x^{m} + a_{1}x^{m+1} + \dots + a_{k}x^{m+k}) /(b_{0}x^{n} + b_{1}x^{n+1} + \dots + b_{l}x^{n+l})$ as $$x$$ tends to $$0$$ by positive or negative values.

[If $$m > n$$, $$\lim\phi(x) = 0$$. If $$m = n$$, $$\lim\phi(x) = a_{0}/b_{0}$$. If $$m < n$$ and $$n – m$$ is even, $$\phi(x) \to +\infty$$ or $$\phi(x) \to -\infty$$ according as $$a_{0}/b_{0} > 0$$ or $$a_{0}/b_{0} < 0$$. If $$m < n$$ and $$n – m$$ is odd, $$\phi(x) \to +\infty$$ as $$x \to +0$$ and $$\phi(x) \to -\infty$$ as $$x \to -0$$, or $$\phi(x) \to -\infty$$ as $$x \to +0$$ and $$\phi(x) \to +\infty$$ as $$x \to -0$$, according as $$a_{0}/b_{0} > 0$$ or $$a_{0}/b_{0} < 0$$.]

6. Orders of smallness. When $$x$$ is small $$x^{2}$$ is very much smaller, $$x^{3}$$ much smaller still, and so on: in other words $\lim_{x\to 0} (x^{2}/x) = 0,\quad \lim_{x\to 0} (x^{3}/x^{2}) = 0,\ \dots.$

Another way of stating the matter is to say that, when $$x$$ tends to $$0$$, $$x^{2}$$, $$x^{3}$$, … all also tend to $$0$$, but $$x^{2}$$ tends to $$0$$ more rapidly than $$x$$, $$x^{3}$$ than $$x^{2}$$, and so on. It is convenient to have some scale by which to measure the rapidity with which a function, whose limit, as $$x$$ tends to $$0$$, is $$0$$, diminishes with $$x$$, and it is natural to take the simple functions $$x$$, $$x^{2}$$, $$x^{3}$$, … as the measures of our scale.

We say, therefore, that $$\phi(x)$$ is of the first order of smallness if $$\phi(x)/x$$ tends to a limit other than $$0$$ as $$x$$ tends to $$0$$. Thus $$2x + 3x^{2} + x^{7}$$ is of the first order of smallness, since $$\lim(2x + 3x^{2} + x^{7})/x = 2$$.

Similarly we define the second, third, fourth, … orders of smallness. It must not be imagined that this scale of orders of smallness is in any way complete. If it were complete, then every function $$\phi(x)$$ which tends to zero with $$x$$ would be of either the first or second or some higher order of smallness. This is obviously not the case. For example $$\phi(x) = x^{7/5}$$ tends to zero more rapidly than $$x$$ and less rapidly than $$x^{2}$$.

The reader may not unnaturally think that our scale might be made complete by including in it fractional orders of smallness. Thus we might say that $$x^{7/5}$$ was of the $$\frac{7}{5}$$th order of smallness. We shall however see later on that such a scale of orders would still be altogether incomplete. And as a matter of fact the integral orders of smallness defined above are so much more important in applications than any others that it is hardly necessary to attempt to make our definitions more precise.

Orders of greatness. Similar definitions are at once suggested to meet the case in which $$\phi(x)$$ is large (positively or negatively) when $$x$$ is small. We shall say that $$\phi(x)$$ is of the $$k$$th order of greatness when $$x$$ is small if $$\phi(x)/x^{-k} = x^{k}\phi(x)$$ tends to a limit different from $$0$$ as $$x$$ tends to $$0$$.

These definitions have reference to the case in which $$x \to 0$$. There are of course corresponding definitions relating to the cases in which $$x \to \infty$$ or $$x \to a$$. Thus if $$x^{k}\phi(x)$$ tends to a limit other than zero, as $$x \to \infty$$, then we say that $$\phi(x)$$ is of the $$k$$th order of smallness when $$x$$ is large: while if $$(x – a)^{k}\phi(x)$$ tends to a limit other than zero, as $$x \to a$$, then we say that $$\phi(x)$$ is of the $$k$$th order of greatness when $$x$$ is nearly equal to $$a$$.

7.3 $$\lim\sqrt{1 + x} = \lim\sqrt{1 – x} = 1$$. [Put $$1 + x = y$$ or $$1 – x = y$$, and use Ex. XXXV. 8.]

8. $$\lim\{\sqrt{1 + x} – \sqrt{1 – x}\}/x = 1$$. [Multiply numerator and denominator by $$\sqrt{1 + x} + \sqrt{1 – x}$$.]

9. Consider the behaviour of $$\{\sqrt{1 + x^{m}} – \sqrt{1 – x^{m}}\}/x^{n}$$ as $$x \to 0$$, $$m$$ and $$n$$ being positive integers.

10. $$\lim\{\sqrt{1 + x + x^{2}} – 1\}/x = \frac{1}{2}$$.

11. $$\lim\dfrac{\sqrt{1 + x} – \sqrt{1 + x^{2}}}{\sqrt{1 – x^{2}} – \sqrt{1 – x}} = 1$$.

12. Draw a graph of the function $y = \biggl\{\frac{1}{x – 1} + \frac{1}{x – \tfrac{1}{2}} + \frac{1}{x – \tfrac{1}{3}} + \frac{1}{x – \tfrac{1}{4}}\biggr\} \bigg/ \biggl\{\frac{1}{x – 1} + \frac{1}{x – \tfrac{1}{2}} + \frac{1}{x – \tfrac{1}{3}} + \frac{1}{x – \tfrac{1}{4}}\biggr\}.$

Has it a limit as $$x \to 0$$? [Here $$y = 1$$ except for $$x = 1$$, $$\frac{1}{2}$$, $$\frac{1}{3}$$, $$\frac{1}{4}$$, when $$y$$ is not defined, and $$y \to 1$$ as $$x \to 0$$.]

13. $$\lim\dfrac{\sin x}{x} = 1$$.

[It may be deduced from the definitions of the trigonometrical ratios4 that if $$x$$ is positive and less than $$\frac{1}{2}\pi$$ then $\sin x < x < \tan x$ or $\cos x < \frac{\sin x}{x} < 1$ or $0 < 1 – \frac{\sin x}{x} < 1 – \cos x = 2\sin^{2} \tfrac{1}{2} x.$

But $$2\sin^{2} \frac{1}{2} x < 2(\frac{1}{2} x)^{2} {=} \frac{1}{2} x^{2}$$ Hence $$\lim\limits_{x \to +0} \left(1 – \dfrac{\sin x}{x}\right) = 0$$, and $$\lim\limits_{x \to +0} \dfrac{\sin x}{x} = 1$$. As $$\dfrac{\sin x}{x}$$ is an even function, the result follows.]

14. $$\lim \dfrac{1 – \cos x}{x^{2}} = \frac{1}{2}$$.

15. $$\lim \dfrac{\sin \alpha x}{x} = \alpha$$. Is this true if $$\alpha = 0$$?

16. $$\lim \dfrac{\arcsin x}{x} = 1$$. [Put $$x = \sin y$$.]

17. $$\lim \dfrac{\tan \alpha x}{x}= \alpha$$,$$\lim\dfrac{\arctan \alpha x}{x} = \alpha$$.

18. $$\lim \dfrac{\csc x – \cot x}{x} = \frac{1}{2}$$.

19. $$\lim\limits_{x \to 1} \dfrac{1 + \cos \pi x}{\tan^{2}\pi x} = \frac{1}{2}$$.

20. How do the functions $$\sin(1/x)$$, $$(1/x)\sin(1/x)$$, $$x\sin(1/x)$$ behave as $$x \to 0$$? [The first oscillates finitely, the second infinitely, the third tends to the limit $$0$$. None is defined when $$x = 0$$. See . 6, 7, 8.]

21. Does the function $y = \biggl(\sin \frac{1}{x}\biggr)\bigg/\biggr(\sin \frac{1}{x}\biggr)$ tend to a limit as $$x$$ tends to $$0$$? [No. The function is equal to $$1$$ except when $$\sin(1/x) = 0$$; when $$x = 1/\pi$$, $$1/2\pi$$, …, $$-1/\pi$$, $$-1/2\pi$$, …. For these values the formula for $$y$$ assumes the meaningless form $$0/0$$, and $$y$$ is therefore not defined for an infinity of values of $$x$$ near $$x = 0$$.]

22. Prove that if $$m$$ is any integer then $$[x] \to m$$ and $$x – [x] \to 0$$ as $$x \to m+0$$, and $$[x] \to m – 1$$, $$x – [x] \to 1$$ as $$x \to m-0$$.

1. For some further discussion of the notion of a function bounded in an interval see § 102.↩︎
2. Thus in Def. A of SecNo 93 we make a statement about values of $$y$$ such that $$0 < y \leq y_{0}$$, the first of these inequalities being inserted expressly in order to exclude the value $$y = 0$$.↩︎
3. In the examples which follow it is to be assumed that limits as $$x \to 0$$ are required, unless (as in Exs. 19, 22) the contrary is explicitly stated.↩︎
4. The proofs of the inequalities which are used here depend on certain properties of the area of a sector of a circle which are usually taken as geometrically intuitive; for example, that the area of the sector is greater than that of the triangle inscribed in the sector. The justification of these assumptions must be postponed to Ch. VII.↩︎