210 
CALCULUS 
Proof. The function * 
*(«)=w®)- i’i«)]- [/(*)-/(<*)] 
satisfies all the conditions of Rolle’s Theorem, Introduction, p. 430, 
and hence its derivative, 
<№) = 
f(P) -/(a) 
F(ò) - F (a) 
F'(x)-f'(x), 
must vanish for a value of x within the interval. Hence 
f(b)-f(a) 
Fib)- F (a) 
F\X)-f\X) = 0, 
a < X < b. 
By hypothesis, F\x) is never 0 in the interval. Consequently we 
are justified in dividing through by F\X), and thus (9) is estab 
lished. 
The Limit jj, Concluded. 
We can now deduce a more general rule 
for determining the limit of the function (3). Applying (9) to an 
arbitrary sub-interval (a, x), when a < x < b, and remembering that 
f(a)= 0 and F(a)= 0, we see that 
(10) 
f(x) = f{X) 
F(x) F'(Xy 
a < X < x, 
where now we have the same X in numerator and denominator. 
When x approaches a, X will also approach a. Hence, if f'(x) /F\x) 
approaches a limit, f\X)/F\X) will approach the same limit, and 
so will its equal, f(x)/F(x). Thus we have: 
(I) 
lim 
x=a 
F(x) X±a F\x) 
If, then, it turns out on differentiating that /'(«) = 0 and F'(a)= 0, 
we can differentiate again, and so on. 
Example. 
V x — sin x ,. 1 — cos x -, • sin x 
lim : lim = lim 
x=M) X z z=M) 3 X 2 *=y) 6 X 
1 
6* 
* We can divide by F(b)— F (a), since 
F(6)- F(a) = (b - a)F'(Y), 
and neither factor on the right is 0. 
a<Y <b,