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

112. Evaluation of the indeterminate form ${\tfrac {\infty }{\infty }}$. In order to find when

	$\lim _{x\to a}{\frac {f(x)}{F(x)}}$
when	$\lim _{x\to a}f(x)=\inf {\text{and}}\lim _{x\to a}F(x)=\infty ,$
that is, when for x = a the function
	${\frac {f(x)}{F(x)}}$
assumes the indeterminate form
	${\frac {\infty }{\infty }}$

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.

Illustrative Example 1. Evaluate ${\tfrac {\log x}{\csc x}}$ for x = 0.

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

113. Evaluation of the indeterminate form $0\cdot \infty$. 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.