Calculus: Analytic Geometry and Calculus, with Vectors

332 Functions, graphs, and numbers

Choose x2 such that x2 > x, and g(x) > g(xi) when x > x2. It then follows

from (7) that

(8)

O(x) - O(xt) e

g(x)^2 (x > x2).

Choose x3 such that x3 > X2 and j(k(x,)/g(x)j < e/2 when x > x3. Then

(9) I O(x)

g(x)

ct(x) - 0(xi) 0(xi)

g(x) + g(x)

fi(x) - -P (X1)

g(x) H

when x > x3. Therefore,

(10) lim r

Lg(z) -L]= 1' mg z) = 0,

q(x!) < e

g(x)

and this gives the required conclusion (3) which involves "indeterminate forms

of the form ao / oo." The following rather simple examples show how the theorem

is applied.

(11) lim x = lim 1 = 0

xti,e x2+1 X_. 2x

(12) lim^3

x" = lim z x = lim?= 0

X- X+ x+1 x-..3z +1 X_

.T-

(13) lim x+ x = lim 1+ ix = 1

X-.. x+1 x-

(14) lim

log x= lim X-1

= lim - = 0 (p > 0)

X-+. X3, x-x___1 x_, pxP

(15) lim t log t = limflog 1= lim -log x= lim i/x= 0.

a--.W X x x-. W x x-. ao

Remark: Limits of functions of other types can be found by using the above

formulas. For example, to find lim xx, we put y = xx and find that log y =

x--0+

x log x, so lim log y = 0, lim y = 1, and lim xx = 1. Similar arguments

x-.0+ x-.0+ x-.0+

show that lim xhIx = 1.

17 Supposing that f, g, h are three functions satisfying the hypotheses of the

mean-value theorem, show that the function F defined by the first of the formulas

F(x) =

f(x)g(x) h(x)

f(a)g(a) h(a)

f(b) g(b) h(b)

I

f (x*) g'(x*) h'(x*)

f(a) g(a) h(a)

f(b) g(b) h(b)

=0

satisfies the hypotheses of the Rolle theorem and hence that there is a number

x* for which a < x < b and the second formula holds. Examine the case in

which h(x) = 1 for each x.

18 This problem provides an opportunity to learn some very interesting

mathematics but, like a bicycle rider who lacks appreciation of basic principles

of physics and engineering, we can pedal along without it. The following theorem

is a fundamental theorem of the calculus which is stronger than Theorem 4.37

because it does not require that f be continuous.