Rule for evaluating the indeterminate form $\frac {0}{0}$

Differentiate the numerator for a new numerator and the denominator for a new denominator^13.5The value of this new fraction for the assigned value^13.6of 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.

Example 13.9.6   Evaluate $\frac{f(x)}{F(x)} = \frac{x^3 - 3x + 2}{x^3 - x^2 - x - 1}$ when $x=1$.

Solution.

$$\frac{f(1)}{F(1)} = \left. \frac{x^3 - 3x + 2}{x^3 - x^2 + 1} \right]_{x = 1} = \frac{1 - 3 + 2}{1 - 1 - 1 + 1} \ \ \lq\lq = \frac{0}{0}''.$$

Therefore, this is an indeterminate form.

$$\frac{f'(1)}{F'(1)} = \left. \frac{3x^2 - 3}{3x^2 - 2x - 1} \right]_{x = 1} = \frac{3 - 3}{3 - 2 - 1}\ \ \lq\lq = \frac{0}{0}''.$$

Therefore, this is an indeterminate form.

$$\frac{f''(1)}{F''(1)} = \left. \frac{6x}{6x - 2} \right]_{x = 1} = \frac{6}{6 - 2} = \frac{3}{2}. \ \ {\rm Ans.}$$

Example 13.9.7   Evaluate $\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x}$.

Solution.

$$\frac{f(0)}{F(0} = \left. \frac{e^x - e^{-x} - 2x}{x - \sin x} \right]_{x = 0} = \frac{1 - 1 - 0}{0 - 0}\ \ \lq\lq = \frac{0}{0}''.$$

Therefore, this is an indeterminate form.

$$\frac{f'(0)}{F'(0)} = \left. \frac{e^x - e^{-x} - 2}{1 - \cos x} \right]_{x = 0} = \frac{1 + 1 - 2}{1 - 1}\ \ \lq\lq = \frac{0}{0}''.$$

Therefore, this is an indeterminate form.

$$\frac{f''(0)}{F''(0)} = \left. \frac{e^x - e^{-x}}{\sin x} \right]_{x = 0} = \frac{1 - 1}{0}\ \ \lq\lq = \frac{0}{0}''.$$

Therefore, this is an indeterminate form.

$$\frac{f'''(0)}{F'''(0)} = \left. \frac{e^x - e^{-x}}{\cos x} \right]_{x = 0} = \frac{1 + 1}{1} = 2. \ \ {\rm Ans.}$$