Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/196

Applying the Theorem of Mean Value to each of these functions (replacing b by x), we get

$f(x)=f(a)+(x-a)f'(x_{1}),$	$a<x_{l}<x$
$F(x)=F(a)+(x-a)F'(x_{2}).$	$a<x_{2}<x$

Since $f(a)=0$ and $F(a)=0$, we get, after canceling out (x - a),

${\frac {f(x)}{F(x)}}={\frac {f'(x_{1})}{F'(x_{2})}}.$

Now let $x{\dot {=}}a$; then $x_{l}{\dot {=}}a,x_{2}{\dot {=}}a$, and

$\lim _{x\to a}f'(x_{1})=f'(a),\lim _{x\to a}F'(x_{2})=F'(a)$.

(49)

∴ $\lim _{x\to a}{\frac {f(x)}{F(x)}}={\frac {f'(a)}{F'(a)}}.$

$F'(a)\neq 0$

Rule for evaluating the indeterminate form ${\tfrac {0}{0}}$. Differentiate the numerator for a new numerator and the denominator for a new denominator.^[1] The value of this new fraction for the assigned value^[2] of the variable will be the limiting value of the original fraction.

In case it so happens that

$f'(a)=0$ and $F'(a)=0$,

that is, the first derivatives also vanish for x = a, then we still have the indeterminate form ${\frac {0}{0}}$, and the theorem can be applied anew to the ratio

${\frac {f'(x)}{F'(x)}}$

giving us

$\lim _{x\to a}{\frac {f(x)}{F(x)}}={\frac {f''(a)}{F''(a)}}.$ When also $f''(a)=0$ and $F''(a)=0$, we get in the same manner

$\lim _{x\to a}{\frac {f(x)}{F(x)}}={\frac {f'''(a)}{F'''(a)}},$

and so on.

It may be necessary to repeat this process several times.

1. The student is warned against the very careless but common mistake of differentiating the whole expression as a fraction by VII.
2. If $a=\inf$, the substitution $x={\frac {1}{z}}$ reduces the problem to the evaluation of the limit for z = 0. Thus $\lim _{x\to \inf }{\frac {f(x)}{F(x)}}=\lim _{z\to 0}{\frac {-f'\left({\frac {1}{z}}\right){\frac {1}{z^{2}}}}{-F'\left({\frac {1}{z}}\right){\frac {1}{z^{2}}}}}=\lim _{z\to 0}{\frac {f'\left({\frac {1}{z}}\right)}{F'\left({\frac {1}{z}}\right)}}=\lim _{x\to \inf }{\frac {f'(x)}{F'(x)}}.$ Therefore the rule holds in this case also.