Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/197

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




 * Solution.
 * $$\frac{f(1)}{F(1)} = \left. \frac{x^3 - 3x + 2}{x^3 - x^2 + 1} \right]_{x = 1} = \frac{1 - 3 + 2}{1 - 1 - 1 + 1} = \frac{0}{0}.$$ &there4; indeterminate.
 * $$\frac{f'(1)}{F'(1)} = \left. \frac{3x^2 - 3}{3x^2 - 2x - 1} \right]_{x = 1} = \frac{3 - 3}{3 - 2 - 1} = \frac{0}{0}.$$ &there4; indeterminate.
 * $$\frac{f(1)}{F(1)} = \left. \frac{6x}{6x - 2} \right]_{x = 1} = \frac{6}{6 - 2} = \frac{3}{2}.$$ Ans.
 * }
 * $$\frac{f(1)}{F(1)} = \left. \frac{6x}{6x - 2} \right]_{x = 1} = \frac{6}{6 - 2} = \frac{3}{2}.$$ Ans.
 * }
 * $$\frac{f(1)}{F(1)} = \left. \frac{6x}{6x - 2} \right]_{x = 1} = \frac{6}{6 - 2} = \frac{3}{2}.$$ Ans.
 * }

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




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

EXAMPLES

Evaluate the following by differentiation: