1549901369-Elements_of_Real_Analysis__Denlinger_

6.6 *L'Hopital's Rule 351

which can be transformed algebraically into

f (y) f(x)

-----

g(y) g(y)

1 - g(x)

g(y)

M+l. (19)

Since this is true for ally between x 0 and x, we may consider what happens

when y ---+ xt. Since g(y) ---+ +oo as y ---+ xt, we will have 1 - :~: ~ > 0 as

y ---+ xt. So inequality (19) is equivalent to

f(y) > (M + l) [ 1 _ g(x)] + f(x).

g(y) g(y) g(y)

(20)

Now, as y ---+ xt, g(y) ---+ +oo, so the right member of inequality (20)
approaches (M + 1) [1 -OJ + 0 = M + l. Thus, Ve > 0, :3 61 > 0 such that
0 < 61 < 6 and

[

g(x)] f(x)

xo < y < xo + 61 :::::> (M + 1) 1 - g(y) + g(y) > (M + 1) - e. (21)

Since this is true when Ve > 0, it is true when e = l. Thus, :3 61 > 0 such that
0 < 61 < 6 and

Xo < y < Xo + 61 ::::;. (M + 1) [ 1 - :~: ~] + ~~~? > M. (22)

Putting together (20) and (22),

f (y)

xo < y < xo + 61 :::::> g(y) > M.

. f(y).. f(x). f'(x)
Therefore, hm -(-) = +oo = L. That ls, hm -( -) = hm --.,--( ) = L.
y->xci g Y x->xci g X x->a g X

Case 3: a = xci, L = -oo, and lim g(x) = + oo. (Exercise 7)

x-+xci

Cases 4, 5, and 6: a =xi), L = a (finite) real number, +oo, or -oo, and
lim g(x) = +oo. (Exercise 8)
x-+xQ

Cases 7, 8, and 9: a = x 0 , L =a (finite) real number, +oo, or -oo, and
lim g(x) = +oo. (Exercise 9)
x-+xo