Calculus: Analytic Geometry and Calculus, with Vectors

9.2 Derivatives and integrals of exponentialsand logarithms 495

Theorem 9.271 If p is a constant positiveexponent, then the first of

the formulas

(9.272) lim xP = aP, lim xP = aP

holds when a > 0 and the second holds when a 0.
To prove this theorem, let p be a given positive constant and let f
be the function for which f(x) = xP when x > 0. That f is continuous
at a when a > 0 is a consequence of the fact (see Theorem 9.27) that
f' (a) exists when a > 0. This proves that the formulas (9.272)are valid
when a > 0. Since OP = 0, it remains to be proved that

(9.273) lim xP = 0.

®-.o+

With the aid of the facts that

(9.274) lim log x = - oo,

x-.O+

this follows from the equality

(9.275)

lim eh = 0,

h-.-.

xP = eP lo; x.

We conclude with some remarks that lead to a formula which has not
appeared in our work but which nevertheless has much more than his-
torical interest. If we know the basic properties of logarithms, we can
let L be the function for which L(x) = log x when x > 0 and obtain the
formula

(9 28)

L(x + h) - L(x)=log (x + h) - log x= 1

log 1 -f-

h

h h h x

lx

=xhlog 1+x

xlogl+

/h

x/

which is of interest from more than one point of view. If we can show,
without reference to our previous results, that there is a number e for
which
h
(9.281) lim 1 +

h

h = e,

h- o

xx

then we can adopt this new number e for the base of our logarithms and
use (9.281) and the fact that L is continuous and L(e) = 1 to obtain
another proof of the formula L'(x) = 1/x. This formula then provides
a back-handed proof of the formula deex/dx = ex. The difficulty in this
view of (9.28) lies in the fact that "direct" proofs of existence of the
limit in (9.281) are neither brief nor quickly comprehended. A proof
is given at the end of the problems of this section. We now look again
at (9.28) and realize that we have proved that L'(x) = 1/x and hence
that the members of (9.28) must converge to 1/x as h -+ 0. Since L