Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/47

As $\scriptstyle {x\doteq 0}$ from the left, $\scriptstyle {y}$ decreases and approaches $\scriptstyle {e}$ as a limit. As $\scriptstyle {x\doteq 0}$ from the right, $\scriptstyle {y}$ increases and also approaches $\scriptstyle {e}$ as a limit.

As $\scriptstyle {x\doteq \infty }$, $\scriptstyle {y}$ approaches the limit $\scriptstyle {1}$; and as $\scriptstyle {\doteq -1}$ from the right, $\scriptstyle {y}$ increases without limit.

In Chap. XVIII, Ex. 15, p. 233, we will show how to calculate the value of $\scriptstyle {e}$ to any number of decimal places.

Natural logarithms are those which have the number $\scriptstyle {e}$ for base. These logarithms play a very important role in mathematics. When the base is not indicated explicitly, the base $\scriptstyle {e}$ is always understood in what follows in this book. Thus $\scriptstyle {\log _{e}v}$ is written simply $\scriptstyle {\log v}$.

Natural logarithms possess the following characteristic property: If $\scriptstyle {x\doteq 0}$ in any way whatever,$\scriptstyle {\operatorname {limit\;} {\frac {\log(1+x)}{x}}=\operatorname {limit\;} \log(1+x)^{\frac {1}{x}}=\log e=1.}$

24. Expressions assuming the form $\scriptstyle {\frac {\infty }{\infty }}$. As $\scriptstyle {\infty }$ is not a number, the expression $\scriptstyle {\infty \div \infty }$ is indeterminate. To evaluate a fraction assuming this form, the numerator and denominator being algebraic functions, we shall find useful the following

Rule. Divide both numerator and denominator by the highest power of the variable occurring in either. Then substitute the value of the variable.

Illustrative Example 1. Evaluate $\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;{\frac {2x^{3}-3x^{2}+4}{5x-x^{2}-7x^{3}}}}$.

Solution. Substituting directly, we get $\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;{\frac {2x^{3}-3x^{2}+4}{5x-x^{2}-7x^{3}}}={\frac {\infty }{\infty }}}$, which is indeterminate. Hence, following the above rule, we divide both numerator and denominator by $\scriptstyle {x^{3}}$. Then$\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;{\frac {2x^{3}-3x^{2}+4}{5x-x^{2}-7x^{3}}}={\underset {x=\infty }{\operatorname {limit} }}\;{\frac {2-{\frac {3}{x}}+{\frac {4}{x^{3}}}}{{\frac {5}{x^{2}}}-{\frac {1}{x}}-7}}=-{\frac {2}{7}}.\quad Ans.}$

EXAMPLES

Prove the following:

1. $\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\left({\frac {x+1}{x}}\right)=1}$.

Proof.

${\begin{aligned}\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\left({\frac {x+1}{x}}\right)}&\scriptscriptstyle {={\underset {x=\infty }{\operatorname {limit} }}\left(1+{\frac {1}{x}}\right)}\\&\scriptscriptstyle {={\underset {x=\infty }{\operatorname {limit} }}(1)+{\underset {x=\infty }{\operatorname {limit} }}\left({\frac {1}{x}}\right)}\\&\scriptscriptstyle {=1+0=1.}\end{aligned}}$

Th. I, p. 18