Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/199

 Evaluate $$\sec 3x \cos 5x$$ for $$x = \tfrac{\pi}{2}$$.


 * Solution. $$\left. \sec 3x \cos 5x \right]_{x = \frac{\pi}{2}} = \infty \cdot 0.$$ &there4; indeterminate.


 * Substituting $$\tfrac{1}{\cos 3x}$$ for $$\sec 3x$$, the function becomes $$\tfrac{\cos 5x}{\cos 3x} = \tfrac{f(x)}{F(x)}$$.


 * {| style="width: 100%;"


 * style="text-align: right;"|$$\frac{ f(\frac{\pi}{2}) }{ F(\frac{\pi}{2}) }$$
 * $$= \left. \frac{\cos 5x}{\cos 3x} \right]_{x = \frac{\pi}{s}} = \frac{0}{0}.$$ indeterminate.
 * style="text-align: right;"|$$\frac{ f'(\frac{\pi}{2}) }{ F'(\frac{\pi}{2}) }$$
 * $$= \left. \frac{-\cos x}{- \sin x} \right]_{x = \frac{\pi}{s}} = \frac{0}{-1} = 0.$$ Ans.
 * }
 * }

It is possible in general to transform the expression into a fraction which will assume either the form $$\tfrac{0}{0}$$ or $$\tfrac{\infty}{\infty}$$.

Evaluate $$\sec x - \tan x$$ for $$x = \tfrac{\pi}{2}$$.


 * Solution. $$\left. \sec x - \tan x \right]_{x= \tfrac{\pi}{2}} = \infty - \infty.$$ &there4; indeterminate.


 * By Trigonometry, $$\sec x - \tan x = \tfrac{1}{\cos x} - \tfrac{\sin x}{\cos x} = \tfrac{1 - \sin x}{\cos x} = \tfrac{f(x)}{F(x)}$$.


 * {| style="width: 100%;"


 * style="text-align: right;"|$$\frac{ f(\frac{\pi}{2}) }{ F(\frac{\pi}{2}) }$$
 * $$= \left. \frac{1 - \sin x}{\cos x} \right]_{x = \frac{\pi}{2}} = \frac{1 - 1}{0} = \frac{0}{0}.$$ &there4; indeterminate.
 * style="text-align: right;"|$$\frac{ f'(\frac{\pi}{2}) }{ F'(\frac{\pi}{2}) }$$
 * $$= \left. \frac{-\cos x}{-\sin x} \right]_{x = \frac{\pi}{2}} = \frac{0}{-1} = 0.$$ Ans.
 * }
 * }

Evaluate the following expressions by differentiation: