James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A52
appendix G The Logarithm Defined as an Integral
In order to graph y − ln x, we first determine its limits:
4 (a) lim
x l `
ln x − `
(b) lim ln x − 2`
xl0 1
Proof
(a) Using Law 3 with x − 2 and r − n (where n is any positive integer), we have
lns2 n d − n ln 2. Now ln 2 . 0, so this shows that lns2 n d l ` as n l `. But ln x is an
increasing function since its derivative 1yx . 0. Therefore ln x l ` as x l `.
(b) If we let t − 1yx, then t l ` as x l 0 1 . Thus, using (a), we have
y
0 1
y=ln x
x
y
1
lim ln x − lim
x l01 S ln 1 − lim s2ln td − 2`
t l ` tD t l `
If y − ln x, x . 0, then
0 1 e
dy
dx − 1 x
x . 0 and d 2 y
dx − 2 1 2 x , 0
y=ln x
2
■
FIGURE 4
y
1
which shows that ln x is increasing and concave downward on s0, `d. Putting this information
together with (4), we draw the graph of y − ln x in Figure 4.
Since ln 1 − 0 and ln x is an increasing continuous function that takes on arbitrarily
large values, the Intermediate Value Theorem shows that there is a number where ln x
takes on the value 1. (See Figure 5.) This important number is denoted by e.
0
1 e
x
5 Definition e is the number such that ln e − 1.
FIGURE 5
y=ln x
We will show (in Theorem 19) that this definition is consistent with our previous definition
of e.
The Natural Exponential Function
Since ln is an increasing function, it is one-to-one and therefore has an inverse function,
which we denote by exp. Thus, according to the definition of an inverse function,
f 21 sxd − y &? f syd − x
6 expsxd − y &? ln y − x
and the cancellation equations are
f 21 s f sxdd − x
f s f 21 sxdd − x
7 expsln xd − x and lnsexp xd − x
In particular, we have
exps0d − 1 since ln 1 − 0
exps1d − e since ln e − 1
We obtain the graph of y − exp x by reflecting the graph of y − ln x about the line
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