1549901369-Elements_of_Real_Analysis__Denlinger_

424 Chapter 7 • The Riemann Integral

In calculus courses we usually justify the existence of such a number by ap-
pealing to the intermediate value theorem. Since lnx is continuous on (0, +oo),
and lim ln x = - oo, and lim ln x = +oo, we conclude from the intermedi-
x-.o+ x->+oo
ate value theorem that there must be a real number x such that ln x = l. To
justify the uniqueness of such a number we note that ln x is strictly increasing,
so must be 1-1.
However, we cannot take this approach here. That is because we have
already defined e by a different procedure, in Definition 2.5.10. Thus, we need
to take this value of e and prove that ln e = 1. That is what we do next.

Theorem 7.7.5 lne = 1, where e = n--+oo lim (1 + .!.)n n from Definition 2.5.10.

Proof. By definition of e, lne = ln [J~~ (1 + ~r]. The function lnx

is continuous on (0 , + oo), hence is continuous ate. The sequence { (1 + ~t}

converges to e. Thus, the sequential criterion for continuity implies that

ln (1 + ~r--+ lne. That is ,

lne = lim ln (1 + .!.)n

n--+oo n

= lim n ln (1 + .!.)

n-+oo n

ln(l+l)

= lim n.

n->oo .!.

Now, by L'Hopital's rule,

ln(l+l)

X--+00 lim l "'

x

n

x d (1)

lim x+l. ClX x

X->00 _Q__ dx (.!.) x

1

. x
= lill --
x->oo X + 1

= l.

(38)

Thus, by the sequential criterion for limits of functions at oo (See Exercise
4.4-B.8.)

. ln(l+~)
n--+oo hm -^1 = 1. (39)
n
Plugging the result (39) into (38), we have lne = l. •

Corollary 7. 7.6 (a) 'r/r E Q, ln(er) = r;

(b) lim ln x = +oo;

x--++oo

(c) lim lnx = - oo;

x-.o+