108
II - Convergence: Discrete variables
where the dummy variables m and n are independent of each other. From
this form, the sophism - it is one - stands out clearly: it is to assume that
given a double sequence u(m, n) whose terms depend on two integers, in this
case
u(m,n) = n{[(l +x/m)m]l/n -I},
one is entitled to confuse
lim [lim u(m, n)]
n-+oo m--+oo
with
lim u(n, n).
n--+oo
But if one chooses u(m, n) = m/(m + n), one has limm u(m, n) = 1 for all n,
so that limn [limm u(m, n)] = 1, while limn u(n, n) = 1/2.
A more reasonable method of proving (14) uses the fact - which we have
not yet proved - that
exp(x) = lim(l + x/n)n.
Since the function log x is differentiable and so continuous, one has, by the
relation (8.2) at the end of nO 8,
logexp(x)
lim log [(1 + x/nt] = lim n.log(l + x/n) =
I . 10g(1 + x/n) - log 1
X.lm
/
;
xn
since x/n tends to 0, the quotient tends to the derivative of the function log
at x = 1, i.e. to 1, whence one concludes that logexp(x) = x. We shall replace these wishy-washy proofs by correct arguments in Chap. IV and even,
on occasion, earlier (Chap. III, nO 2).
However it may be, these calculations show that it will not be without
value to establish the following result, another example of an application of
Theorem 2:
Example 3. Let p a nonzero integer. Every real number a > 0 possesses one
and only one positive pth root.
In other words, the equation x P = a has a unique root x > O. One may
assume a > 0 and p > 0, since the case where p < 0 reduces to this, by the
identity x- P = l/x p • First, there exists an Xl > 0 such that xi> a since, for
x large, one has x P > x > a. Next, we define a sequence of numbers Xn > 0
by
PX2 = (p - l)XI + a/xi-I,
and, generally,
(10.15)
II - Convergence: Discrete variables
where the dummy variables m and n are independent of each other. From
this form, the sophism - it is one - stands out clearly: it is to assume that
given a double sequence u(m, n) whose terms depend on two integers, in this
case
u(m,n) = n{[(l +x/m)m]l/n -I},
one is entitled to confuse
lim [lim u(m, n)]
n-+oo m--+oo
with
lim u(n, n).
n--+oo
But if one chooses u(m, n) = m/(m + n), one has limm u(m, n) = 1 for all n,
so that limn [limm u(m, n)] = 1, while limn u(n, n) = 1/2.
A more reasonable method of proving (14) uses the fact - which we have
not yet proved - that
exp(x) = lim(l + x/n)n.
Since the function log x is differentiable and so continuous, one has, by the
relation (8.2) at the end of nO 8,
logexp(x)
lim log [(1 + x/nt] = lim n.log(l + x/n) =
I . 10g(1 + x/n) - log 1
X.lm
/
;
xn
since x/n tends to 0, the quotient tends to the derivative of the function log
at x = 1, i.e. to 1, whence one concludes that logexp(x) = x. We shall replace these wishy-washy proofs by correct arguments in Chap. IV and even,
on occasion, earlier (Chap. III, nO 2).
However it may be, these calculations show that it will not be without
value to establish the following result, another example of an application of
Theorem 2:
Example 3. Let p a nonzero integer. Every real number a > 0 possesses one
and only one positive pth root.
In other words, the equation x P = a has a unique root x > O. One may
assume a > 0 and p > 0, since the case where p < 0 reduces to this, by the
identity x- P = l/x p • First, there exists an Xl > 0 such that xi> a since, for
x large, one has x P > x > a. Next, we define a sequence of numbers Xn > 0
by
PX2 = (p - l)XI + a/xi-I,
and, generally,
(10.15)
