1549901369-Elements_of_Real_Analysis__Denlinger_

352 Chapter 6 • Differentiable Functions

Cases 10, 11, and 12: a = +oo, and L = a (finite) real number, +oo, or

-oo, and lim g(x) = +oo. (Exercise 10)

x->+oo

Cases 13, 14, and 15: a= -oo, and L = a (finite) real number, +oo, or

-oo, and li m g(x) = +oo. (Exercise 11)

X-+-cx:>

Cases 16-30: lim g(x) = -oo.
X->Q
Suppose f ,g: I--> JR., where f,g,I, and a satisfy conditions (a)- (e) speci-
fied above, with X->Q lim g(x) = -oo. Define the function h(x) = -g(x). Then

(^1) 1m. -h f I ( ) ( x) = l' 1m --( f I ( x) - ) = -L (fi mte,. + oo or - oo ).
X->Q I X X->Q -g^1 X
Then f and h satisfy hypotheses (a)- (e) above, with X->Q lim h(x) = + oo, and

. f'(x).. f(x)
X--H:k 11m -h I ( ) X = -L (fimte, +oo or - oo). Thus, by Cases 1-^15 , X-+O: hm -h( X ) = -L;
.. f(x). f(x)
that 1s, X->Q hm -(-) g X = -X->Q hm -h( X ) = -(-L) = L. •

Examples 6.6. 7 Calculate each of t he following limits. Before using L'Hopital's
rule, be sure that the hypotheses are met.

(a) lim lnx (b) lim x3 + 4x + 7

x-.oo x x->oo 5x3 - x^2 - 3 ( )

1

. lnsinx
C !ill ---
x->O lnx

Solution: (a) As x--> oo, t he denominator--> oo, so by Theorem 6.6.6,

. lnx. l/x
hm -= hm -=0.
X->00 X X->00 1

(b) As x --> oo, the denominator --> oo, so by Theorem 6.6.6,

1

. x^3 + 4x + 7
1
. 3x^2 + 4
1
. 6x
1m = 1m - 1m ---
x->oo 5x 3 - x^2 - 3 x->oo 15x^2 - 2x x->oo 30x - 2

6 1

30 5

Notice that this answer agrees with the a nswer we would obtain using the

"algebra of limits" of Section 4.2.

(c) As x--> 0, the denominator--> - oo, so by Theorem 6.6.6,

lim^1 n sm. x = lim SinX^1 cos x = lim x cos x.

x-.o ln x x-.o l x x-.o sin x