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

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

$$\lim_{x \to a} f'(x_1) = f'(a), \lim_{x \to a} F'(x_2) = F'(a)$$.

Rule for evaluating the indeterminate form $$\tfrac{0}{0}$$. Differentiate the numerator for a new numerator and the denominator for a new denominator. The value of this new fraction for the assigned value 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.