Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/198



In order to find when

we follow the same rule as that given in §111 for evaluating the indeterminate form $$\tfrac{0}{0}$$. Hence

Rule for evaluating the indeterminate form $$\tfrac{\infty}{\infty}$$. ''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.''

A rigorous proof of this rule is beyond the scope of this book and is left for more advanced treatises.

Evaluate $$\tfrac{\log x}{\csc x}$$ for x = 0.




 * Solution.
 * $$\frac{f(0)}{F(0)} = \left. \frac{\log(x)}{\csc(x)} \right]_{x = 0} = \frac{-\infty}{\infty}.$$ &there4; indeterminate.
 * $$\frac{f'(0)}{F'(0)} = \left. \frac{\frac{1}{x}}{-\csc x \cot x} \right]_{x = 0} = \left. -\frac{\sin^2 x}{x \cos x} \right]_{x = 0} = \frac{0}{0}.$$ &there4; indeterminate.
 * $$\frac{f(0)}{F(0)} = \left. -\frac{2 \sin x \cos x}{\cos x - x \sin x} \right]_{x = 0} = -\frac{0}{1} = 0.$$ Ans.
 * }
 * $$\frac{f(0)}{F(0)} = \left. -\frac{2 \sin x \cos x}{\cos x - x \sin x} \right]_{x = 0} = -\frac{0}{1} = 0.$$ Ans.
 * }
 * $$\frac{f(0)}{F(0)} = \left. -\frac{2 \sin x \cos x}{\cos x - x \sin x} \right]_{x = 0} = -\frac{0}{1} = 0.$$ Ans.
 * }

If a function $$f(x) \cdot \phi(x)$$ takes on the indeterminate form $$0 \cdot \infty$$ for $$x = a$$, we write the given function

$$f(x) \cdot \phi(x) = \frac{f(x)}{ \frac{1}{\phi(x)} }$$ $$\left( \text{or} = \frac{\phi(x)}{\frac{1}{f(x)}} \right)$$

so as to cause it to take on one of the forms $$\tfrac{0}{0}$$ or $$\tfrac{\infty}{\infty}$$, thus bringing it under §111 or §112.