Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/201


 * {| style="width:100%"


 * By § 112,
 * style="text-align: right;"|$$\ \log y$$
 * $$\frac{\frac{1}{x}}{-\frac{1}{x^2}} = -x = 0,$$
 * when $$x = 0$$.
 * }


 * Since $$y = x^x$$, this gives $$log_e x^x = 0$$; i.e., $$x^x = 1$$. Ans.

Evaluate $$(1 + x)^{\tfrac{1}{x}}$$; when $$x = 0$$.


 * Solution. This function assumes the indeterminate form $$1^{\infty}$$ for $$x = 0$$.


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


 * Let
 * style="text-align: right;"|$$\ y$$
 * colspan="2"|$$= (1 + x)^{\frac{1}{x}};$$
 * then
 * style="text-align: right;"|$$\ \log y$$
 * $$= \frac{1}{x} \log(1 + x) = \infty \cdot . 0,$$
 * when $$x = 0 $$.
 * By § 113,
 * style="text-align: right;"|$$\ y$$
 * $$= \frac{\log(1 + x)}{x} = \frac{0}{0},$$
 * when x = 0.
 * By § 111,
 * style="text-align: right;"|$$\ y$$
 * $$= \frac{\frac{1}{1 + x}}{1} = \frac{1}{1 + x} = 1,$$
 * when $$x = 0 $$.
 * }
 * $$= \frac{\frac{1}{1 + x}}{1} = \frac{1}{1 + x} = 1,$$
 * when $$x = 0 $$.
 * }


 * since $$y = (1 + x)^{\frac{1}{1 + x}}$$, this gives $$log_e (1 + x)^{\frac{1}{x}} = 1$$; i.e. $$(1 + x)^{\frac{1}{x}} = e$$. Ans.

Evaluate $$(\cot x)^{\sin x}$$ for $$x = 0$$.


 * Solution. This function assumes the indeterminate form $$\infty^0$$ for $$x = 0$$.


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


 * Let
 * style="text-align: right;"|$$\ y$$
 * colspan="2"|$$=\ (\cot x)^{\sin x};$$
 * then
 * style="text-align: right;"|$$\ \log y$$
 * $$=\ \sin x \log \cot x = 0 \cdot \infty,$$
 * when $$x = 0$$.
 * By § 113,
 * style="text-align: right;"|$$\ \log y$$
 * $$= \frac{\log \cot x}{\csc x} = \frac{\infty}{\infty},$$
 * when $$x = 0$$.
 * By § 112,
 * style="text-align: right;"|$$\ \log y$$
 * $$= \frac{\frac{-\csc^2 x}{\cot x}}{-\csc x \cot x} = \frac{\sin x}{\cos^2 x} = 0,$$
 * when x = 0.
 * }
 * $$= \frac{\frac{-\csc^2 x}{\cot x}}{-\csc x \cot x} = \frac{\sin x}{\cos^2 x} = 0,$$
 * when x = 0.
 * }

EXAMPLES

Evaluate the following expressions by differentiation: