the derivative of which
is infinite at x = 0.A6.1 Example of a Function which cannot be expanded in a Taylor Series
The possibility of expressing a function by
means of a Taylor series with a remainder of the (n +
l)-th order depends essentially on the differentiability of the
function at the point in question. For this reason, the function
log x cannot be represented into a Taylor series in
powers of x and can the function the derivative of which
is infinite at x = 0.
In fact, for a function to be capable of expansion in an infinite Taylor series all its derivatives must necessarily exist at the point in question; however, this condition is by no means sufficient. A function for which all the derivatives exist and are continuous throughout an interval still need not necessarily be capable of expansion in a Taylor series, i.e., the remainder Rn in Taylor's theorem may fail to tend to zero as n increases, no matter how small is the interval in which we wish to expand the function.
The simplest example of this phenomenon is offered by the function y=f(x)=e-1/x² for x ¹ 0, f(0) = 0, which we have already encountered in A3.1.1. This function, with all its derivatives, is continuous in every interval, even at x=0, and we have seen that at this point all the derivatives vanish, i.e., that f(n)(0) = 0 for every value of n. Hence all the coefficients of the polynomials of approximation vanish in Taylor's theorem, no matter what is the value chosen for n. In other words, the remainder is and remains equal to the function itself and therefore, except when x = 0, does not approach 0 as n increases, since the function is positive for every non-zero value of x.
A6.2. Proof that e
is Irrational:
We immediately deduce from the
formula 
that the number e is irrational. In fact,
if the contrary were true, i.e., if e = p/q,
where p and q are integers, we can certainly
choose n larger than q. Then n!e=n!p/q
must be an integer. On the other hand,
and, since eq < e < 3, we must have 0 < eq/(n + 1) < 1, whence the integer
n!e = 2n! + n!/2 + ··· 1 plus a non-vanishing proper fraction,
which is impossible.
A6.3 Proof that the Binomial Series Converges: In 6.3.3, we have postponed the estimation of the remainder Rn in the expansion of f(x)=(1+x)a for |x| < 1. We shall now carry out this estimate. It is most convenient to separate the cases x > 0 and x < 0.
For f (n+1)(x), we have the expression
If x > 0, we write the remainder in Lagrange's form
so that
Letting b = [|a|] + 1, where [|a|] denotes the greatest integer which does not exceed |a|, we have
and since b is fixed, if 0 < x < 1, this approaches 0 as n increases.
For the case -1 < x < 0, we write the remainder in Cauchy's form
so that
Since |x| < 1, the last factor cannot exceed a constant K, independent of n. Moreover, (l - q)/(l - q |x|) < l. As before, writing b = [|a|] + 1, we have
which approaches 0 as n increases.
Thus, in either case when |x| < 1, the remainder tends to zero as n increases, justifying the expansion in 6.3.3.
A6.4 Zeroes and Infinities of Functions and So-called Intermediate Expressions: The Tailor series for a function in the neighbourhood of a point x=a enables us to characterize the behaviour of the function in the neighbourhood of this point as follows: We say that a function f(x) at x = a has an exactly n-tuple zero or it vanishes there exactly to the order n, if f(a) = 0, f '(a) = 0, f "(a) = 0, ··· , f (n-1)(a) = 0, and f (n)(a) ¹ 0. We assume here that the function possesses in the neighbourhood of the point at least continuous derivatives to the n-th order. By our definition, we seek to indicate that the Taylor series for the function in the neighbourhood of the point can be written in the form
in which, as h ® 0, the factor F(h) tends to a limit different from 0, namely, the value f n(a).
If a function f (x) is defined at all points in the neighbourhood of a point x = a, except perhape at x = a itself, and if
where at the point x = a the numerator does not vanish, but the denominator has a n-tuple zero, we say that the function f (x) becomes infinite at the n-th order at the point x = a. If at the point x = a the numerator also has a m-tuple zero and if m > n, we say that the function has a (m - n)-tuple zero there; while if m < n, we say that the function has a (n - m)-tuple infinity.
All these definitions are in agreement with the convention regarding functional behaviour already laid down (cf. 3.9.3). In order to make these relations precise, we expand the numerator and denominator by Taylor's theorem, using Lagrange's form of the remainder; the function then has the form
in which q and q1 are two numbers between 0 and 1 and the factors by which hm/m! and hn/n! are multiplied do not tend to zero as h does, since they approach the limits +f (m)(a) and g(n)(a), respectively, which differ from zero. If m > n, we have then
Accordingly, the expression f (x) vanishes to the order m - n. If n > mt, we see that the expression f (a + h) becomes infinite at the order n - m as h ® 0. If m=n, we obtain the equation
We can express the content of the last equation as follows: If the numerator and denominator of a function f (x) = f (x)/g(x) both vanish at x = a, we can determine the limiting value as x ® a by differentiating the numerator and denominator an equal number of times until at least one of the derivatives is non-zero. If this happens for the numerator and denominator simultaneously, the limit, which we are seeking, is equal to the quotient of these two derivatives. If we encounter a non-vanishing derivative in the denominator earlier than in the numerator, the fraction tends to zero. If we encounter a non-vanishing derivative in the numerator earlier than in the denominator, the absolute value of the fraction increases beyond all bounds.
We thus have a rule for evaluating the socalled indeterminate expressions 0/0 - a subject that is discussed at exaggerated length in many textbooks on the differential and integral calculus. In reality, the point in question is merely the very simple determination of the limiting value of a quotient in which the numerator and the denominator tend to zero. The name indeterminate expression, usually found in the literature, is misleading and vague.
We can arrive at our results in a somewhat different way by basing the proof on the generalized mean value theorem instead of on Taylor's theorem (A2.2).
Thi method of deriving our rule has the advantage that there is no use made of the existence of the derivative at the point x = a itself; moreover, it includes the case in which f(x) is defined for x ³ a only, so that the passage to the limit x ® a or h ® 0 is made from one side only.
According to this, if g'(x) ¹ 0, we have
where q is the same in both the numerator and denominator. Hence, in particular, when f(a) = 0 = g(a),
where q is a value in the interral 0 <q < 1 and, if we set k = qh, we obtain
it being assumed that the limit on the right hand side exists. If
we can proceed in the same manner until we come to the first index for which it is no longer true that f (m)(a) = 0 = g(m)(a). Then
where we also include the case in which both sides have the limit infinity.
As examples, consider
as x ® 0. We have
Moreover, note that other so-called
indeterminate forms can alao be reduced to the case considered;
for example, the limit of 
as x ® 0, being the limit of the
difference of two expressions both of which beome infinite, is an
indeterminate form ¥ - ¥. By the transformation
we arrive at once at an expression the limit of which as x ® 0 is determined by our rule
Evaluate the limits 1-12:
13. Prove that the function y = (x²)x~, y(0) = 1 is continuous at x = 0.