LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
Excerpts
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
The proof just given is somewhat subtle and indirect, and it may be well, in view of the great importance of the theorem, to indicate alternative lines of proof.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
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 $x = 0$.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
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.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
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 $(0, 1)$.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
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 $0$.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
The only difference between the ‘tending of $n$ to $\infty$’ discussed in the last chapter, and this ‘tending of $x$ to $\infty$’, is that $x$ assumes all values as it tends to $\infty$, *i.e.* that the point $P$ which corresponds to $x$ coincides in turn with every point of $\Lambda$ to the right of its initial position, whereas $n$ tended to $\infty$ by a series of jumps. We can express this distinction by saying that $x$ tends *continuously* to $\infty$.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
It is equal to zero whenever $x$ is an integer, so that the function $\phi(n)$ derived from it is always zero and so tends to the limit zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
The reader should observe carefully that to assert the continuity of $\phi(x, y)$ with respect to the two variables $x$ and $y$ is to assert much more than its continuity with respect to each variable considered separately. It is plain that if $\phi(x, y)$ is continuous with respect to $x$ and $y$ then it is certainly continuous with respect to $x$ (or $y$) when any fixed value is assigned to $y$ (or $x$). But the converse is by no means true.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
We shall call such a square a *neighbourhood* of $(a, b)$, and say that the condition in question is satisfied *in the neighbourhood of $(a, b)$*, or *near $(a, b)$*, meaning by this simply that it is possible to find *some* square throughout which the condition is satisfied.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
This theorem is of fundamental importance in the theory of definite integrals (Ch.VII). It is impossible, without the use of this or some similar theorem, to prove that a function continuous throughout an interval necessarily possesses an integral over that interval.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
If we had supposed that $y^{2} = x$ then the conditions of the theorem would not have been satisfied, for $y^{2}$ is not a steadily increasing function of $y$ in any interval which includes $y = 0$: it decreases when $y$ is negative and increases when $y$ is positive. And in this case the conclusion of the theorem does not hold, for $y^{2} = x$ defines *two* functions of $x$, viz. $y = \sqrt{x}$ and $y = -\sqrt{x}$, both of which vanish when $x = 0$, and each of which is defined only for positive values of $x$, so that the equation has sometimes two solutions and sometimes none.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
There is nothing to show that the $\EPSILON_{1}$ of the conclusion is the $\EPSILON$ of the hypotheses, and indeed this is generally untrue.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
The reader may be tempted to think that this proof is needlessly elaborate, and that the existence of points of the interval, not in any interval of $I$, follows at once from the fact that the sum of all these intervals is less than $1$.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
This class has an upper bound $\eta$, and plainly $F(\eta) \leq \xi$. If $F(\eta)$ were less than $\xi$, we could find a value of $y$ such that $y > \eta$ and $F(y) < \xi$, and $\eta$ would not be the upper bound of the class considered. Hence $F(\eta) = \xi$.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
It is indeed obvious that there are an infinity of such intervals corresponding to every $\xi$ and every $\DELTA$, for if the condition is satisfied for any particular value of $\EPSILON$, then it is satisfied *a fortiori* for any smaller value.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
This statement is apparently simpler; but it contains phrases the precise meaning of which has not yet been explained and can only be explained by the help of inequalities like those which occur in our original statement.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
Another method of stating the definition is this: *$\phi(x, y)$ is continuous for $x = \xi$, $y = \eta$ if $\phi(x, y) \to \phi(\xi, \eta)$ when $x \to \xi$, $y \to \eta$ in any manner*. This statement is apparently simpler; but it contains phrases the precise meaning of which has not yet been explained and can only be explained by the help of inequalities like those which occur in our original statement.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
It is impossible, without the use of this or some similar theorem, to prove that a function continuous throughout an interval necessarily possesses an integral over that interval.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
If $\phi(x) = 1/q$ when $x = p/q$, and $\phi(x) = 0$ when $x$ is irrational, then $\phi(x)$ is continuous for all irrational and discontinuous for all rational values of $x$.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
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 $0$.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
The equation (2) 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 $(0, 1)$.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
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.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
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 (x/x) = 1, for $x/x$ is equal to $1$ so long as $x$ differs from zero, however small the difference may be.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
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 $x^{2}$.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
In fact, if $x = -y$ and $\phi(x) = \phi(-y) = \psi(y)$, then $y$ tends to $\infty$ as $x$ tends to $-\infty$, and the question of the behaviour of $\phi(x)$ as $x$ tends to $-\infty$ is the same as that of the behaviour of $\psi(y)$ as $y$ tends to $\infty$.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
The function $\phi(x) = x - [x]$ oscillates between $0$ and $1$, as is obvious from the form of its graph. It is equal to zero whenever $x$ is an integer, so that the function $\phi(n)$ derived from it is always zero and so tends to the limit zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
It is natural to call a function *continuous* if its graph is a continuous curve, and otherwise discontinuous.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
The function $\phi(x)$ is said to be continuous for $x = \xi$ if it tends to a limit as $x$ tends to $\xi$ from either side, and each of these limits is equal to $\phi(\xi)$.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
Thus our definition asserts that if we draw two such horizontal lines, no matter how close together, we can always cut off a vertical strip of the plane by two vertical lines in such a way that all that part of the curve which is contained in the strip lies between the two horizontal lines.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
An ‘infinity’ is the kind of discontinuity of most common occurrence in ordinary work.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
In other words *as $x$ varies from $x_{0}$ to $x_{1}$, $y$ must assume at least once every value between $y_{0}$ and $y_{1}$*.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
Indeed it is not even true that $\phi(x)$ must be continuous when it assumes each value *once and once only*.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
The net result of this and the last section is consequently to show that our common-sense notion of what we mean by continuity is substantially accurate, and capable of precise statement in mathematical terms.
Equations
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
|\phi(x) - l| < \DELTAFor x at or beyond some x0(Delta), phi(x) differs from l by less than any chosen positive Delta, which is the condition for phi(x) to tend to l as x tends to infinity.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim_{x \to \infty} \phi(x) = lphi(x) tends to the limit l as x tends to infinity.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\phi(x) \to -\inftyphi(x) tends to -infinity with x, the mirror of the definition of tending to +infinity.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\phi(x) \to \inftyNotation for phi(x) tending to +infinity, as in Definition 2.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim_{y \to +0} \phi(y) = lphi(y) tends to the limit l as y tends to 0 through positive values.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim_{y \to 0} \phi(y) = lphi(y) tends to the limit l as y tends to 0 from both sides, with y different from zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim_{x \to a} \phi(x) = lphi(x) tends to the limit l as x tends to a; the value at x = a itself plays no part.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim_{x \to a+0} \phi(x) = lphi(x) tends to l as x approaches a from the right, through values greater than a.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim_{x \to a+0} \phi(x) = \phi(a+0)The right-hand limit of phi at a is written phi(a+0).
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\phi(a-0) \leq \phi(a) \leq \phi(a+0)For a function that is steadily increasing near a, its left-hand limit at a is no greater than its value at a, which is no greater than its right-hand limit.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\phi(x) = 0If phi(x) is identically zero, its limit as x tends to zero is zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\psi(x) = 0A function equal to phi(x) except that psi(0) = 1 still has limit zero at x = 0, since the limit ignores the value at 0.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\psi(x) = [1 - x^{2}]psi(x) is the greatest integer not greater than 1 - x^2; it equals 1 at x = 0 and 0 for 0 < |x| < 1.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim(x/x) = 1The function x/x has limit 1 as x tends to zero, although it is not defined at x = 0.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\phi(x) = 2For phi(x) = ((x+1)^2 - 1)/x, which equals x + 2 for x not zero, the limit as x tends to zero is 2.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\phi(x) = \{(x + 1)^{2} - 1\}/x = x + 2For x not zero, phi(x) equals x + 2, so it is defined by the expression except at x = 0.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\dfrac{\sin x}{x} = 1The ratio of sine x to x tends to 1 as x tends to zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\sin x < x < \tan xFor x positive and less than pi/2, sine x is less than x, which is less than tangent x; this is the inequality used to prove the limit of sin x over x.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim \dfrac{1 - \cos x}{x^{2}} = \frac{1}{2}The quantity 1 minus cosine x, divided by x squared, tends to one half as x tends to zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim \dfrac{\arcsin x}{x} = 1The ratio of the inverse sine of x to x tends to 1 as x tends to zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim \dfrac{\sin \alpha x}{x} = \alphaThe ratio of sin(alpha x) to x tends to alpha as x tends to zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim \dfrac{\tan \alpha x}{x}= \alphaThe ratio of tan(alpha x) to x tends to alpha as x tends to zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim \dfrac{\cosec x - \cot x}{x} = \frac{1}{2}The difference cosecant x minus cotangent x, divided by x, tends to one half as x tends to zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\limits_{x \to 1} \dfrac{1 + \cos \pi x}{\tan^{2}\pi x} = \frac{1}{2}The ratio of 1 + cos(pi x) to tan^2(pi x) tends to one half as x tends to 1.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\limits_{x \to a} (x^{2} - a^{2})/(x - a) = 2aThe difference quotient of x squared tends to 2a as x tends to a.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\limits_{x \to a} (x^{m} - a^{m})/(x - a) = ma^{m-1}For any integer m, the difference quotient of x^m tends to m a^(m-1) as x tends to a.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\limits_{x \to 1} (x^{7} - 2x^{5} + 1)/(x^{3} - 3x^{2} + 2) = 1The ratio of the two polynomials tends to 1 as x tends to 1, since x - 1 is a factor of both.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\limits_{x \to a} x^{m} = a^{m}For any integer m (except a = 0 with m negative), x^m tends to a^m as x tends to a.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\limits_{x \to a} R(x) = R(a)A rational function R tends to its value at a whenever a is not a root of its denominator.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\limits_{x \to 0} (a + bx + cx^{2} + \dots + kx^{m}) = aA polynomial in x tends to its constant term as x tends to zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\limits_{x \to 0} \left\{(a + bx + \dots + kx^{m})/(\alpha + \beta x + \dots + \kappa x^{\mu})\right\} = a/\alphaThe ratio of two polynomials in x tends to a over alpha as x tends to zero, provided alpha is not zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\phi(x) + \psi(x) \to l + l'The sum of two functions tending to limits l and l' tends to l + l'.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\phi(x)\psi(x) \to ll'The product of two functions tending to limits l and l' tends to ll'.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\phi(x)/\psi(x) \to l/l'The quotient of two functions tending to limits l and l' tends to l/l', unless l' is zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim_{x\to 0} (x^{2}/x) = 0x^2 is of smaller order than x as x tends to zero, since x^2/x tends to zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\phi(x)/x^{-k} = x^{k}\phi(x)Dividing phi(x) by x^(-k) is the same as multiplying it by x^k; the order of greatness of phi when x is small is decided by this product tending to a nonzero limit.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\sqrtp{1 + x} = \lim\sqrtp{1 - x} = 1The square root of 1 + x and the square root of 1 - x both tend to 1 as x tends to zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\{\sqrtp{1 + x} - \sqrtp{1 - x}\}/x = 1The difference of the two square roots divided by x tends to 1 as x tends to zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\{\sqrtp{1 + x + x^{2}} - 1\}/x = \frac{1}{2}The square root of 1 + x + x^2, less 1, divided by x tends to one half as x tends to zero.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
H < \phi(x) < Kphi(x) is bounded in an interval around a when it lies between two fixed constants H and K there.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lambda = \Lambda = lphi(x) tends to l as x tends to a exactly when its lower and upper limits of indetermination both equal l.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
|\phi(x_{2}) - \phi(x_{1})| < \DELTAThe principle of convergence: phi(x) tends to a limit as x tends to a exactly when values of phi at two points close to a differ by less than any given Delta.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\phi(x_{2}) \geq \phi(x_{1})phi(x) is steadily increasing with x when its value at any larger x_2 is at least its value at any smaller x_1.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\phi(x_{2}) > \phi(x_{1})phi(x) is steadily increasing in the stricter sense when, for x_2 > x_1, its value at x_2 is strictly greater than at x_1.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
|\phi(x) - \phi(\xi)| < \DELTAThe value of the function at x stays within a given small amount DELTA of its value at xi; this is the fundamental inequality of continuity, which the book shows is equivalent to the limit definition.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\phi(\xi - 0) = \phi(\xi) = \phi(\xi + 0)A function is continuous at xi exactly when its left-hand limit, its value at xi, and its right-hand limit at xi are all equal.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\phi(a - 0) = \phi(a) = \phi(a + 0)For continuity at x = a, the one-sided limits from below and above must both exist and equal the value phi(a).
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\phi(x) \leq KThe function is bounded above when some fixed number K is never exceeded by any of its values in the interval.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
O(a, b) = M(a, b) - m(a, b)The oscillation of a bounded function in an interval is the difference between its upper bound and its lower bound in that interval.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
O(a, b) \leq O(a, c) + O(c, b)The oscillation over the whole interval from a to b is at most the sum of the oscillations over the two parts a to c and c to b.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
R(x) = P(x)/Q(x)The standard rational function R(x) is the quotient of two polynomials P(x) and Q(x).
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\sin(x + h) - \sin x = 2\sin \tfrac{1}{2}h \cos(x + \tfrac{1}{2}h)The difference between sin of x plus h and sin of x equals twice sin of half h times cos of x plus half h, which shows sin x is continuous.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
|\phi(x, y) - \phi(\xi, \eta) | < \DELTAThe function phi(x, y) is continuous at (xi, eta) when every value within a small enough square around that point differs from phi(xi, eta) by less than any given positive Delta.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\phi(x, y) = \frac{2xy}{x^{2} + y^{2}}Example function of two variables, defined as 2xy/(x^2+y^2) away from the axes and 0 when x or y is zero, which is continuous in each variable separately but not jointly at (0, 0).
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
\lim\phi(x, y) = \frac{2a}{1 + a^{2}}Along the straight line y = ax, the example function tends to 2a/(1+a^2), which depends on a, so the joint limit at (0, 0) does not exist.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
y^{5} - xy - y - x = 0The relation between x and y that defines y as an implicit function of x, which by the theorem has a unique continuous solution vanishing with x.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
f(x, y) - f(x, y') = (y - y') (y^{4} + y^{3}y' + y^{2}y'^{2} + yy'^{3} + y'^{4} - x - 1)For f(x, y) = y^5 - xy - y - x, the difference f(x, y) - f(x, y') factors as (y - y') times a second factor, which is the basis for checking that f is steadily decreasing in y.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
y = \tfrac{1}{2}\{1 + x - \sqrtp{1 + 6 x + x^{2}}\}Solution of the quadratic y^2 - xy - y - x = 0 as an explicit function of x, using the positive square root.
LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. CONTINUOUS AND DISCONTINUOUS FUNCTIONS
f(a, \lambda) = 0Because f{x, phi(x)} = 0 and f is continuous, the value of f at the limit point (a, lambda) is zero, which forces lambda = b.
Problems
Exercise XXXVII
Exercise XXXVII, problem 1, p. 176
The sum or product of two functions continuous at a point is continuous at that point. The quotient is also continuous unless the denominator vanishes at the point. [This follows at once from % [examples:xxxv]Ex. xxxv%. 1.]
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 10, p. 176
For what values of $x$ are $\tan x$, $\cot x$, $\sec x$, and $\cosec x$ continuous or discontinuous?
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 11, p. 176
If $f(y)$ is continuous for $y = \eta$, and $\phi(x)$ is a continuous function of $x$ which is equal to $\eta$ when $x = \xi$, then $f\{\phi(x)\}$ is continuous for $x = \xi$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 12, p. 176
If $\phi(x)$ is continuous for any particular value of $x$, then any polynomial in $\phi(x)$, such as $a\{\phi(x)\}^{m} + \dots$, is so too.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 13, p. 176
Discuss the continuity of 1/(a^2 x + b^2 x),0pt minus 3pt2 + x,0pt minus 3pt1 + x,0pt minus 3pt1/1 + x.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 14, p. 176
$\sin(1/x)$, $x\sin(1/x)$, and $x^{2}\sin(1/x)$ are continuous except for $x = 0$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 15, p. 176
The function which is equal to $x\sin(1/x)$ except when $x = 0$, and to zero when $x = 0$, is continuous for all values of $x$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 16, p. 176
$[x]$ and $x - [x]$ are discontinuous for all integral values of $x$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 17, p. 176
For what (if any) values of $x$ are the following functions discontinuous: $[x^{2}]$, $[\sqrt{x}\,]$, $\sqrtp{x - [x]}$, $[x] + \sqrtp{x - [x]}$, $[2x]$, $[x] + [-x]$?
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 18, p. 176
**of discontinuities.** Some of the preceding examples suggest a classification of different types of discontinuity. 0pt minus 3pt% [2.25em][l](1)% [2.25em][l](1)% % Suppose that $\phi(x)$ tends to a limit as $x \to a$ either by values less than or by values greater than $a$. Denote these limits, as in [§]95, by $\phi(a - 0)$ and $\phi(a + 0)$ respectively. Then, for continuity, it is necessary and sufficient that $\phi(x)$ should be defined for $x = a$, and that $\phi(a - 0) = \phi(a) = \phi(a + 0)$. Discontinuity may arise in a variety of ways. [1.5em][l]($\alpha$) $\phi(a - 0)$ may be equal to $\phi(a + 0)$, but $\phi(a)$ may not be defined, or may differ from $\phi(a - 0)$ and $\phi(a + 0)$. Thus if $\phi(x) = x \sin(1/x)$ and $a = 0$, $\phi(0 - 0) = \phi(0 + 0) = 0$, but $\phi(x)$ is not defined for $x = 0$. Or if $\phi(x) = [1 - x^{2}]$ and $a = 0$, $\phi(0 - 0) = \phi(0 + 0) = 0$, but $\phi(0) = 1$. [1.5em][l]($\beta$) 0.375em plus 0.75em minus 0.25em$\phi(a - 0)$ and $\phi(a + 0)$ may be unequal. In this case $\phi(a)$ may be equal to one or to neither, or be undefined. The first case is illustrated by $\phi(x) = [x]$, for which $\phi(0 - 0) = -1$, $\phi(0 + 0) = \phi(0) = 0$; the second by $\phi(x) = [x] - [-x]$, for which $\phi(0 - 0) = -1$, $\phi(0 + 0) = 1$, $\phi(0) = 0$; and the third by $\phi(x) = [x] + x \sin(1/x)$, for which $\phi(0 - 0)= -1$, $\phi(0 + 0) = 0$, and $\phi(0)$ is undefined. In any of these cases we say that $\phi(x)$ has a **discontinuity** at $x = a$. And to these cases we may add those in which $\phi(x)$ is defined only on one side of $x = a$, and $\phi(a - 0)$ or $\phi(a + 0)$, as the case may be, exists, but $\phi(x)$ is either not defined when $x = a$ or has when $x = a$ a value different from $\phi(a - 0)$ or $\phi(a + 0)$. It is plain from [§]95 that *a function which increases or decreases steadily in the neighbourhood of $x = a$ can have at most a simple discontinuity for $x = a$*. 0pt minus 3pt% [2.25em][l](2)% [2.25em][l](2)% % It may be the case that only one (or neither) of $\phi(a - 0)$ and $\phi(a + 0)$ exists, but that, supposing for example $\phi(a + 0)$ not to exist, $\phi(x) \to +\infty$ or $\phi(x) \to -\infty$ as $x \to a+0$, so that $\phi(x)$ tends to a limit or to $+\infty$ or to $-\infty$ as $x$ approaches $a$ from either side. Such is the case, for instance, if $\phi(x) = 1/x$ or $\phi(x) = 1/x^{2}$, and $a = 0$. In such cases we say (cf. Ex. 7) that $x = a$ is an **** of $\phi(x)$. And again we may add to these cases those in which $\phi(x) \to +\infty$ or $\phi(x) \to -\infty$ as $x \to a$ from one side, but $\phi(x)$ is not defined at all on the other side of $x = a$. 0pt minus 3pt% [2.25em][l](3)% [2.25em][l](3)% % Any point of discontinuity which is not a point of simple discontinuity nor an infinity is called a point of **discontinuity**. Such is the point $x = 0$ for the functions $\sin(1/x)$, $(1/x)\sin(1/x)$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 19, p. 176
What is the nature of the discontinuities at $x = 0$ of the functions $(\sin x)/x$, $[x] + [-x]$, $\cosec x$, $\sqrtp{1/x}$, $\sqrtp[3]{1/x}$, $\cosec(1/x)$, $\sin(1/x)/\sin(1/x)$?
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 2, p. 176
Any polynomial is continuous for all values of $x$. Any rational fraction is continuous except for values of $x$ for which the denominator vanishes. [This follows from xxxv. 6, 7.]
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 20, p. 176
The function which is equal to $1$ when $x$ is rational and to $0$ when $x$ is irrational (Ch.II, % [examples:xvi]Ex. xvi%. 10) is discontinuous for all values of $x$. So too is any function which is defined only for rational or for irrational values of $x$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 21, p. 176
0.375em plus 0.75em minus 0.25emThe function which is equal to $x$ when $x$ is irrational and to $\sqrtb{(1 + p^{2})/(1 + q^{2})}$ when $x$ is a rational fraction $p/q$ (Ch.II, % [examples:xvi]Ex. xvi%. 11) is discontinuous for all negative and for positive rational values of $x$, but continuous for positive irrational values.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 22, p. 176
For what points are the functions considered in Ch.IV, xxxi discontinuous, and what is the nature of their discontinuities? [Consider, *e.g.*, the function $y = \lim x^{n}$ (Ex. 5). Here $y$ is only defined when $-1 < x \leq 1$: it is equal to $0$ when $-1 < x < 1$ and to $1$ when $x = 1$. The points $x = 1$ and $x = -1$ are points of simple discontinuity.]
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 3, p. 176
$\sqrt{x}$ is continuous for all positive values of $x$ (% [examples:xxxv]Ex. xxxv%. 8). It is not defined when $x < 0$, but is continuous for $x = 0$ in virtue of the remark made at the end of [§]98. The same is true of $x^{m/n}$, where $m$ and $n$ are any positive integers of which $n$ is even.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 4, p. 176
The function $x^{m/n}$, where $n$ is odd, is continuous for all values of $x$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 5, p. 176
$1/x$ is not continuous for $x = 0$. It has no value for $x = 0$, nor does it tend to a limit as $x \to 0$. In fact $1/x \to +\infty$ or $1/x \to -\infty$ according as $x \to 0$ by positive or negative values.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 6, p. 176
Discuss the continuity of $x^{-m/n}$, where $m$ and $n$ are positive integers, for $x = 0$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 7, p. 176
The standard rational function $R(x) = P(x)/Q(x)$ is discontinuous for $x = a$, where $a$ is any root of $Q(x) = 0$. Thus $(x^{2} + 1)/(x^{2} - 3x + 2)$ is discontinuous for $x = 1$. It will be noticed that in the case of rational functions a discontinuity is always associated with (*a*) a failure of the definition for a particular value of $x$ and (*b*) a tending of the function to $+\infty$ or $-\infty$ as $x$ approaches this value from either side. Such a particular kind of point of discontinuity is usually described as an **** of the function. An ‘infinity’ is the kind of discontinuity of most common occurrence in ordinary work.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 8, p. 176
Discuss the continuity of (x - a)(b - x),0pt minus 3pt[3](x - a)(b - x),0pt minus 3pt(x - a)/(b - x),0pt minus 3pt[3](x - a)/(b - x)
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVII, problem 9, p. 176
$\sin x$ and $\cos x$ are continuous for all values of $x$. [We have (x + h) - x = 212h (x + 12h), which is numerically less than the numerical value of $h$.]
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVIII
Exercise XXXVIII, problem 1, p. 184
0.375em plus 0.75em minus 0.25emIf $\phi(x) = 1/x$ except when $x = 0$, and $\phi(x) = 0$ when $x = 0$, then $\phi(x)$ has neither an upper nor a lower bound in any interval which includes $x = 0$ in its interior, as *e.g.* the interval $\DPmod{(-1, +1)}{[-1, +1]}$.
Printed answer:- 0.375em plus 0.75em minus 0.25emIf $\phi(x) = 1/x$ except when $x = 0$, and $\phi(x) = 0$ when $x = 0$, then $\phi(x)$ has neither an upper nor a lower bound in any interval which includes $x = 0$ in its interior, as *e.g.* the interval $\DPmod{(-1, +1)}{[-1, +1]}$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVIII, problem 2, p. 184
If $\phi(x) = 1/x^{2}$ except when $x = 0$, and $\phi(x) = 0$ when $x = 0$, then $\phi(x)$ has the lower bound $0$, but no upper bound, in the interval $\DPmod{(-1, +1)}{[-1, +1]}$.
Printed answer:- If $\phi(x) = 1/x^{2}$ except when $x = 0$, and $\phi(x) = 0$ when $x = 0$, then $\phi(x)$ has the lower bound $0$, but no upper bound, in the interval $\DPmod{(-1, +1)}{[-1, +1]}$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVIII, problem 3, p. 184
Let $\phi(x) = \sin(1/x)$ except when $x = 0$, and $\phi(x) = 0$ when $x = 0$. Then $\phi(x)$ is discontinuous for $x = 0$. In any interval $\DPmod{(-\DELTA, +\DELTA)}{[-\DELTA, +\DELTA]}$ the lower bound is $-1$ and the upper bound $+1$, and each of these values is assumed by $\phi(x)$ an infinity of times.
Printed answer:- Let $\phi(x) = \sin(1/x)$ except when $x = 0$, and $\phi(x) = 0$ when $x = 0$. Then $\phi(x)$ is discontinuous for $x = 0$. In any interval $\DPmod{(-\DELTA, +\DELTA)}{[-\DELTA, +\DELTA]}$ the lower bound is $-1$ and the upper bound $+1$, and each of these values is assumed by $\phi(x)$ an infinity of times.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVIII, problem 4, p. 184
Let $\phi(x) = x - [x]$. This function is discontinuous for all integral values of $x$. In the interval $\DPmod{(0, 1)}{[0, 1]}$ its lower bound is $0$ and its upper bound $1$. It is equal to $0$ when $x = 0$ or $x = 1$, but it is never equal to $1$. Thus $\phi(x)$ never assumes a value equal to its upper bound.
Printed answer:- Let $\phi(x) = x - [x]$. This function is discontinuous for all integral values of $x$. In the interval $\DPmod{(0, 1)}{[0, 1]}$ its lower bound is $0$ and its upper bound $1$. It is equal to $0$ when $x = 0$ or $x = 1$, but it is never equal to $1$. Thus $\phi(x)$ never assumes a value equal to its upper bound.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVIII, problem 5, p. 184
Let $\phi(x) = 0$ when $x$ is irrational, and $\phi(x) = q$ when $x$ is a rational fraction $p/q$. Then $\phi(x)$ has the lower bound $0$, but no upper bound, in any interval $\DPmod{(a, b)}{[a, b]}$. But if $\phi(x) = (-1)^{p}q$ when $x = p/q$, then $\phi(x)$ has neither an upper nor a lower bound in any interval.
Printed answer:- Let $\phi(x) = 0$ when $x$ is irrational, and $\phi(x) = q$ when $x$ is a rational fraction $p/q$. Then $\phi(x)$ has the lower bound $0$, but no upper bound, in any interval $\DPmod{(a, b)}{[a, b]}$. But if $\phi(x) = (-1)^{p}q$ when $x = p/q$, then $\phi(x)$ has neither an upper nor a lower bound in any interval.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V
Exercise Misc-V, problem 1, p. 194
Show that, if neither $a$ nor $b$ is zero, then ax^n + bx^n-1 + …+ k = ax^n (1 + _x), where $\epsilon_{x}$ is of the first order of smallness when $x$ is large.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 10, p. 194
Prove that $\phi(x) = 1 - \cos(1 - \cos x)$ is of the fourth order of smallness when $x$ is small; and find the limit of $\phi(x)/x^{4}$ as $x \to 0$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 11, p. 194
Prove that $\phi(x) = x\sin(\sin x) - \sin^{2}x$ is of the sixth order of smallness when $x$ is small; and find the limit of $\phi(x)/x^{6}$ as $x \to 0$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 12, p. 194
From a point $P$ on a radius $OA$ of a circle, produced beyond the circle, a tangent $PT$ is drawn to the circle, touching it in $T$, and $TN$ is drawn perpendicular to $OA$. Show that $NA/AP \to 1$ as $P$ moves up to $A$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 13, p. 194
Tangents are drawn to a circular arc at its middle point and its extremities; $\Delta$ is the area of the triangle formed by the chord of the arc and the two tangents at the extremities, and $\Delta'$ the area of that formed by the three tangents. Show that $\Delta/\Delta' \to 4$ as the length of the arc tends to zero.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 14, p. 194
For what values of $a$ does $\{a + \sin(1/x)\}/x$ tend to (1) $\infty$, (2) $-\infty$, as $x \to 0$? [To $\infty$ if $a > 1$, to $-\infty$ if $a < -1$: the function oscillates if $-1 \leq a \leq 1$.]
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 15, p. 194
If $\phi(x) = 1/q$ when $x = p/q$, and $\phi(x) = 0$ when $x$ is irrational, then $\phi(x)$ is continuous for all irrational and discontinuous for all rational values of $x$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 16, p. 194
Show that the function whose graph is drawn in [fig:32]Fig. 32 may be represented by either of the formulae 1 - x + [x] - [1 - x],0pt minus 3pt1 - x - _n (^2n+1x).
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 17, p. 194
Show that the function $\phi(x)$ which is equal to $0$ when $x = 0$, to $\frac{1}{2} - x$ when $0 < x < \frac{1}{2}$, to $\frac{1}{2}$ when $x = \frac{1}{2}$, to $\frac{3}{2} - x$ when $\frac{1}{2}< x < 1$, and to $1$ when $x = 1$, assumes every value between $0$ and $1$ once and once only as $x$ increases from $0$ to $1$, but is discontinuous for $x = 0$, $x = \frac{1}{2}$, and $x = 1$. Show also that the function may be represented by the formula 12 - x - 12[2x] - 12[1 - 2x].
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 18, p. 194
Let $\phi(x) = x$ when $x$ is rational and $\phi(x) = 1 - x$ when $x$ is irrational. Show that $\phi(x)$ assumes every value between $0$ and $1$ once and once only as $x$ increases from $0$ to $1$, but is discontinuous for every value of $x$ except $x = \frac{1}{2}$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 19, p. 194
As $x$ increases from $-\frac{1}{2}\pi$ to $\frac{1}{2}\pi$, $y = \sin x$ is continuous and steadily increases, in the stricter sense, from $-1$ to $1$. Deduce the existence of a function $x = \arcsin y$ which is a continuous and steadily increasing function of $y$ from $y = -1$ to $y = 1$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 2, p. 194
If $P(x) = ax^{n} + bx^{n-1} + \dots + k$, and $a$ is not zero, then as $x$ increases $P(x)$ has ultimately the sign of $a$; and so has $P(x + \lambda) - P(x)$, where $\lambda$ is any constant.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 20, p. 194
Show that the numerically least value of $\arctan y$ is continuous for all values of $y$ and increases steadily from $-\frac{1}{2}\pi$ to $\frac{1}{2}\pi$ as $y$ varies through all real values.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 21, p. 194
Discuss, on the lines of [§§]108--109, the solution of the equations y^2 - y - x = 0,0pt minus 3pty^4 - y^2 - x^2 = 0,0pt minus 3pty^4 - y^2 + x^2 = 0 in the neighbourhood of $x = 0$, $y = 0$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 22, p. 194
If $ax^{2} + 2bxy + cy^{2} + 2dx + 2ey = 0$ and $\Delta = 2bde - ae^{2} - cd^{2}$, then one value of $y$ is given by $y = \alpha x + \beta x^{2} + (\gamma + \epsilon_{x}) x^{3}$, where = -d/e,0pt minus 3pt = /2e^3,0pt minus 3pt= (cd - be) /2e^5, and $\DPtypo{e_{x}}{\epsilon_{x}}$ is of the first order of smallness when $x$ is small. [If $y - \alpha x = \eta $ then -2e = ax^2 + 2bx(+ x) + c(+ x)^2 = Ax^2 + 2Bx + C^2, say. It is evident that $\eta$ is of the second order of smallness, $x\eta$ of the third, and $\eta^{2}$ of the fourth; and $-2e\eta = Ax^{2} - (AB/e) x^{3}$, the error being of the fourth order.]
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 23, p. 194
If $x = ay + by^{2} + cy^{3}$ then one value of $y$ is given by y = x + x^2 + (+ _x) x^3, where $\alpha = 1/a$, $\beta = -b/a^{3}$, $\gamma = (2b^{2} - ac)/a^{5}$, and $\epsilon_{x}$ is of the first order of smallness when $x$ is small.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 24, p. 194
If $x = ay + by^{n}$, where $n$ is an integer greater than unity, then one value of $y$ is given by $y = \alpha x + \beta x^{n} + (\gamma + \epsilon_{x}) x^{2n-1}$, where $\alpha = 1/a$, $\beta = -b/a^{n+1}$, $\gamma = nb^{2}/a^{2n+1}$, and $\epsilon_{x}$ is of the $(n - 1)$th order of smallness when $x$ is small.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 25, p. 194
Show that the least positive root of the equation $xy = \sin x$ is a continuous function of $y$ throughout the interval $\DPmod{(0, 1)}{[0, 1]}$, and decreases steadily from $\pi$ to $0$ as $y$ increases from $0$ to $1$. [The function is the inverse of $(\sin x)/x$: apply [§]109.]
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 26, p. 194
The least positive root of $xy = \tan x$ is a continuous function of $y$ throughout the interval $\DPmod{(1, \infty)}{[1, \infty)}$, and increases steadily from $0$ to $\frac{1}{2}\pi$ as $y$ increases from $1$ towards $\infty$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 3, p. 194
Show that in general (ax^n + bx^n-1 + …+ k)/(Ax^n + Bx^n-1 + …+ K) = + (/x) (1 + _x), where $\alpha = a/A$, $\beta = (bA - aB)/A^{2}$, and $\epsilon_{x}$ is of the first order of smallness when $x$ is large. Indicate any exceptional cases.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 4, p. 194
Express (ax^2 + bx + c)/(Ax^2 + Bx + C) in the form + (/x) + (/x^2)(1 + _x), where $\epsilon_{x}$ is of the first order of smallness when $x$ is large.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 5, p. 194
Show that _xxx + a - x = 12 a. [Use the formula $\sqrtp{x + a} - \sqrt{x} = a/\{\sqrtp{x + a} + \sqrt{x}\}$.]
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 6, p. 194
Show that $\sqrtp{x + a} = \sqrt{x} + \frac{1}{2}(a/\sqrt{x}) (1 + \epsilon_{x})$, where $\epsilon_{x}$ is of the first order of smallness when $x$ is large.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 7, p. 194
Find values of $\alpha$ and $\beta$ such that $\sqrtp{a x^{2} + 2bx + c} - \alpha x - \beta$ has the limit zero as $x \to \infty$; and prove that $\lim x\{\sqrtp{ax^{2} + 2bx + c} - \alpha x - \beta\} = (ac - b^{2})/2a$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 8, p. 194
Evaluate _x xx^2 + x^4 + 1 - x2.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise Misc-V, problem 9, p. 194
Prove that $(\sec x - \tan x) \to 0$ as $x \to \frac{1}{2}\pi$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXIV
Exercise XXXIV, problem 1, p. 164
Consider the behaviour of the following functions as $x \to \infty$: $1/x$, $1 + (1/x)$, $x^{2}$, $x^{k}$, $[x]$, $x - [x]$, $[x] + \sqrtb{x - [x]}$. The first four functions correspond exactly to functions of $n$ fully discussed in Ch.IV. The graphs of the last three were constructed in Ch.II (xvi. 1, 2, 4), and the reader will see at once that $[x] \to \infty$, $x - [x]$ oscillates finitely, and $[x] + \sqrtb{x - [x]} \to \infty$. One simple remark may be inserted here. The function $\phi(x) = x - [x]$ oscillates between $0$ and $1$, as is obvious from the form of its graph. It is equal to zero whenever $x$ is an integer, so that the function $\phi(n)$ derived from it is always zero and so tends to the limit zero. The same is true if (x) = x,0pt minus 3pt(n) = n= 0. It is evident that $\phi(x) \to l$ or $\phi(x) \to \infty$ or $\phi(x) \to -\infty$ involves the corresponding property for $\phi(n)$, but that the converse is by no means always true.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXIV, problem 2, p. 164
Consider in the same way the functions: (x)/x,0pt minus 3ptxx,0pt minus 3pt(xx)^2,0pt minus 3ptx,0pt minus 3pta^2 x+ b^2 x, illustrating your remarks by means of the graphs of the functions.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXIV, problem 3, p. 164
Give a geometrical explanation of Def. 1, analogous to the geometrical explanation of Ch.IV, [§]59.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXIV, problem 4, p. 164
If $\phi(x) \to l$, and $l$ is not zero, then $\phi(x)\cos x\pi$ and $\phi(x)\sin x\pi$ oscillate finitely. If $\phi(x) \to \infty$ or $\phi(x) \to -\infty$, then they oscillate infinitely. The graph of either function is a wavy curve oscillating between the curves $y = \phi(x)$ and $y = -\phi(x)$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXIV, problem 5, p. 164
Discuss the behaviour, as $x \to \infty$, of the function y = f(x)^2 x+ F(x)^2 x, where $f(x)$ and $F(x)$ are some pair of simple functions (*e.g.* $x$ and $x^{2}$). [The graph of $y$ is a curve oscillating between the curves $y = f(x)$, $y = F(x)$.]
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXV
Exercise XXXV, problem 1, p. 168
If (x) l,0pt minus 3pt(x) 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 $z \to a$. [pg]169 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 $0 < |x - a| \leq \EPSILON$ for $n \geq n_{0}$.]
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXV, problem 2, p. 168
If $m$ is a positive integer then $x^{m} \to 0$ as $x \to 0$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXV, problem 3, p. 168
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 $x^{m} \to 1$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXV, problem 4, p. 168
$\lim\limits_{x \to 0} (a + bx + cx^{2} + \dots + kx^{m}) = a$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXV, problem 5, p. 168
$\lim\limits_{x \to 0} \left\{(a + bx + \dots + kx^{m})/(\alpha + \beta x + \dots + \kappa x^{\mu})\right\} = a/\alpha$, 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 % [examples:xxxvi]Ex. xxxvi%. 5. Discuss the cases which arise when $a \neq 0$ and more than one of the first coefficients in the denominator vanish.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXV, problem 6, p. 168
$\lim\limits_{x \to a} x^{m} = a^{m}$, 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.]
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXV, problem 7, p. 168
$\lim\limits_{x \to a} R(x) = R(a)$, if $R$ denotes any rational function and $a$ is not one of the roots of its denominator.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXV, problem 8, p. 168
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. % [examples:xxviii]Ex. xxviii%. 4). If $a$ is negative we write $x = -y$ and $a = -b$. Then x^m = (-1)^my^m = (-1)^mb^m = a^m.]
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVI
Exercise XXXVI, problem 1, p. 171
$\lim\limits_{x \to a} (x^{2} - a^{2})/(x - a) = 2a$.
Printed answer:- $\lim\limits_{x \to a} (x^{2} - a^{2})/(x - a) = 2a$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles2*a
Exercise XXXVI, problem 10, p. 171
$\lim\{\sqrtp{1 + x + x^{2}} - 1\}/x = \frac{1}{2}$.
Printed answer:- $\lim\{\sqrtp{1 + x + x^{2}} - 1\}/x = \frac{1}{2}$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles1/2
Exercise XXXVI, problem 11, p. 171
$\lim\dfrac{\sqrtp{1 + x} - \sqrtp{1 + x^{2}}}{\sqrtp{1 - x^{2}} - \sqrtp{1 - x}} = 1$.
Printed answer:- $\lim\dfrac{\sqrtp{1 + x} - \sqrtp{1 + x^{2}}}{\sqrtp{1 - x^{2}} - \sqrtp{1 - x}} = 1$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles1
Exercise XXXVI, problem 12, p. 171
Draw a graph of the function y = 1x - 1 + 1x - 12 + 1x - 13 + 1x - 14 / 1x - 1 + 1x - 12 + 1x - 13 + 1x - 14. Has it a limit as $x \to 0$?
Printed answer:- 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 $x \to 0$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVI, problem 13, p. 171
$\lim\dfrac{\sin x}{x} = 1$.
Printed answer:- $\lim\dfrac{\sin x}{x} = 1$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles1
Exercise XXXVI, problem 14, p. 171
$\lim \dfrac{1 - \cos x}{x^{2}} = \frac{1}{2}$.
Printed answer:- $\lim \dfrac{1 - \cos x}{x^{2}} = \frac{1}{2}$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles1/2
Exercise XXXVI, problem 15, p. 171
$\lim \dfrac{\sin \alpha x}{x} = \alpha$. Is this true if $\alpha = 0$?
Printed answer:- $\lim \dfrac{\sin \alpha x}{x} = \alpha$. Is this true if $\alpha = 0$?
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handlesalpha
Exercise XXXVI, problem 16, p. 171
$\lim \dfrac{\arcsin x}{x} = 1$.
Printed answer:- $\lim \dfrac{\arcsin x}{x} = 1$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles1
Exercise XXXVI, problem 17a, p. 171
$\lim \dfrac{\tan \alpha x}{x}= \alpha$,0pt minus 3pt$\lim\dfrac{\arctan \alpha x}{x} = \alpha$.
Printed answer:- $\lim \dfrac{\tan \alpha x}{x}= \alpha$,0pt minus 3pt$\lim\dfrac{\arctan \alpha x}{x} = \alpha$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handlesalpha
Exercise XXXVI, problem 17b, p. 171
$\lim \dfrac{\tan \alpha x}{x}= \alpha$,0pt minus 3pt$\lim\dfrac{\arctan \alpha x}{x} = \alpha$.
Printed answer:- $\lim \dfrac{\tan \alpha x}{x}= \alpha$,0pt minus 3pt$\lim\dfrac{\arctan \alpha x}{x} = \alpha$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handlesalpha
Exercise XXXVI, problem 18, p. 171
$\lim \dfrac{\cosec x - \cot x}{x} = \frac{1}{2}$.
Printed answer:- $\lim \dfrac{\cosec x - \cot x}{x} = \frac{1}{2}$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles1/2
Exercise XXXVI, problem 19, p. 171
$\lim\limits_{x \to 1} \dfrac{1 + \cos \pi x}{\tan^{2}\pi x} = \frac{1}{2}$.
Printed answer:- $\lim\limits_{x \to 1} \dfrac{1 + \cos \pi x}{\tan^{2}\pi x} = \frac{1}{2}$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles1/2
Exercise XXXVI, problem 2, p. 171
$\lim\limits_{x \to a} (x^{m} - a^{m})/(x - a) = ma^{m-1}$, if $m$ is any integer (zero included).
Printed answer:- $\lim\limits_{x \to a} (x^{m} - a^{m})/(x - a) = ma^{m-1}$, if $m$ is any integer (zero included).
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handlesm*a**(m-1)
Exercise XXXVI, problem 20, p. 171
0.375em plus 0.75em minus 0.25emHow do the functions $\sin(1/x)$, $(1/x)\sin(1/x)$, $x\sin(1/x)$ behave as $x \to 0$?
Printed answer:- [The first oscillates finitely, the second infinitely, the third tends to the limit $0$. None is defined when $x = 0$. See xv. 6, 7, 8.]
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVI, problem 21, p. 171
Does the function y = (1x)/(1x) tend to a limit as $x$ tends to $0$?
Printed answer:- *No*. The function is equal to $1$ except when $\sin(1/x) = 0$; *i.e.* 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 $x = 0$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVI, problem 22, p. 171
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 $x \to m-0$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVI, problem 3, p. 171
Show that the result of Ex. 2 remains true for all rational values of $m$, provided $a$ is positive.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVI, problem 4, p. 171
$\lim\limits_{x \to 1} (x^{7} - 2x^{5} + 1)/(x^{3} - 3x^{2} + 2) = 1$.
Printed answer:- $\lim\limits_{x \to 1} (x^{7} - 2x^{5} + 1)/(x^{3} - 3x^{2} + 2) = 1$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles1
Exercise XXXVI, problem 5, p. 171
Discuss the behaviour of (x) = (a_0x^m + a_1x^m+1 + …+ a_kx^m+k) /(b_0x^n + b_1x^n+1 + …+ b_lx^n+l) as $x$ tends to $0$ by positive or negative values.
Printed answer:- 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 $a_{0}/b_{0} < 0$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVI, problem 6, p. 171
**of smallness**. When $x$ is small $x^{2}$ is very much smaller, $x^{3}$ much smaller still, and so on: in other words _x0 (x^2/x) = 0,0pt minus 3pt_x0 (x^3/x^2) = 0, …. 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. **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 $a$.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles
Exercise XXXVI, problem 7, p. 171
$\lim\sqrtp{1 + x} = \lim\sqrtp{1 - x} = 1$.
Printed answer:- $\lim\sqrtp{1 + x} = \lim\sqrtp{1 - x} = 1$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles1
Exercise XXXVI, problem 8, p. 171
$\lim\{\sqrtp{1 + x} - \sqrtp{1 - x}\}/x = 1$.
Printed answer:- $\lim\{\sqrtp{1 + x} - \sqrtp{1 - x}\}/x = 1$.
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles1
Exercise XXXVI, problem 9, p. 171
Consider the behaviour of $\{\sqrtp{1 + x^{m}} - \sqrtp{1 - x^{m}}\}/x^{n}$ as $x \to 0$, $m$ and $n$ being positive integers.
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
other: not a kind the checker handles