1549901369-Elements_of_Real_Analysis__Denlinger_

350 Chapter 6 • Differentiable Functions

Now, as y --+ xci, g(y) --+ +oo, so the left member of this inequality

approaches (L - c) [1 - OJ + 0 = L - c, and the right member approaches

(L + c) [1 -OJ+ 0 = L + c. Thus, 3 6 1 > 0 such that 0 < '51 < 6 and

Xo < y < Xo + '51 =? (L - c) [ 1 - :n + ~? > (L - c) - c

and (L + c) [ 1 - :~] + ~? < (L + c) + c.

Thus, by (16) and these last two inequalities,

f (y)

xo < y < xo + '51 => L - 2c < g(y) < L + 2c

=> lf(y) - LI< 2c.

g(y)

. f(y) f(x)
Therefore, hm -( -) = L. Equivalently, lim -( -) = L.
y->x;j g Y x->x;j g X

Case 2: a= xci, L = +oo, and lim g(x) = +oo.

x~xci

Again, suppose f,g: I--+ IR, where f,g, and I satisfy conditions (b)-(e)

specified above.

Let M > 0.

By hypothesis (e), 3 6 > 0 :7 Xo + 6 EI and

f'(x)

x^0 < x < xo + 6 => g' ( x) > M + l. (17)

Suppose x is any number satisfying x 0 < x < x 0 + 8, and let y be any
number between x 0 and x. As in the proof of Case 1, the Cauchy mean value
theorem applies to f and g on the closed interval [y, x], so 3 Cx,y E (y, x ) such
that

f(x) - f(y) _ f'(cx,y)

g(x) - g(y) - g'(cx,y) ·

y

I

x

I

Figure 6.12

Thus, by (17) and (18), whenever Xo < x < xo + 8,

f(x) - f(y) > M + 1

g(x) - g(y) '

(18)