Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/45

Written in the form of limits.		Abbreviated form often used.
${\begin{array}{rrclrcl}\scriptstyle {{\text{(1)}}~}&\scriptstyle {{\underset {x=0}{\operatorname {limit} }}\;{\frac {c}{x}}}&\scriptstyle {=}&\scriptstyle {\infty ;}&\scriptstyle {\frac {c}{0}}&\scriptstyle {=}&\scriptstyle {\infty .}\\\scriptstyle {{\text{(2)}}~}&\scriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;cx}&\scriptstyle {=}&\scriptstyle {\infty ;}&\scriptstyle {c\cdot \infty }&\scriptstyle {=}&\scriptstyle {\infty .}\\\scriptstyle {{\text{(3)}}~}&\scriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;{\frac {x}{c}}}&\scriptstyle {=}&\scriptstyle {\infty ;}&\scriptstyle {\frac {\infty }{c}}&\scriptstyle {=}&\scriptstyle {\infty .}\\\scriptstyle {{\text{(4)}}~}&\scriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;{\frac {c}{x}}}&\scriptstyle {=}&\scriptstyle {0;}&\scriptstyle {\frac {c}{\infty }}&\scriptstyle {=}&\scriptstyle {0.}\\\scriptstyle {{\text{(5)}}~}&\scriptstyle {{\underset {x=-\infty }{\operatorname {limit} }}\;a^{x}}&\scriptstyle {=}&\scriptstyle {+\infty ,{\text{ when }}a<1;}&\scriptstyle {a^{-\infty }}&\scriptstyle {=}&\scriptstyle {+\infty .}\\\scriptstyle {{\text{(6)}}~}&\scriptstyle {{\underset {x=+\infty }{\operatorname {limit} }}\;a^{x}}&\scriptstyle {=}&\scriptstyle {0,\quad {\text{ when }}a<1;}&\scriptstyle {a^{+\infty }}&\scriptstyle {=}&\scriptstyle {0.}\\\scriptstyle {{\text{(7)}}~}&\scriptstyle {{\underset {x=-\infty }{\operatorname {limit} }}\;x^{x}}&\scriptstyle {=}&\scriptstyle {0,\quad {\text{ when }}a>1;}&\scriptstyle {a^{-\infty }}&\scriptstyle {=}&\scriptstyle {0.}\\\scriptstyle {{\text{(8)}}~}&\scriptstyle {{\underset {x=+\infty }{\operatorname {limit} }}\;a^{x}}&\scriptstyle {=}&\scriptstyle {+\infty {\text{ when}}a>1;}&\scriptstyle {a^{+\infty }}&\scriptstyle {=}&\scriptstyle {+\infty .}\\\scriptstyle {{\text{(9)}}~}&\scriptstyle {{\underset {x=0}{\operatorname {limit} }}\;\log _{a}x}&\scriptstyle {=}&\scriptstyle {+\infty {\text{ when }}a<1;}&\scriptstyle {\log _{a}0}&\scriptstyle {=}&\scriptstyle {+\infty .}\\\scriptstyle {{\text{(10)}}~}&\scriptstyle {{\underset {x=+\infty }{\operatorname {limit} }}\;\log _{a}x}&\scriptstyle {=}&\scriptstyle {-\infty ,{\text{ when}}a<1;}&\scriptstyle {\log _{a}(+\infty )}&\scriptstyle {=}&\scriptstyle {-\infty .}\\\scriptstyle {{\text{(11)}}~}&\scriptstyle {{\underset {x=0}{\operatorname {limit} }}\;\log _{a}x}&\scriptstyle {=}&\scriptstyle {-\infty ,{\text{ when }}a>1;}&\scriptstyle {\log _{a}0}&\scriptstyle {=}&\scriptstyle {-\infty .}\\\scriptstyle {{\text{(12)}}~}&\scriptstyle {{\underset {x=+\infty }{\operatorname {limit} }}\;\log _{a}x}&\scriptstyle {=}&\scriptstyle {+\infty ,{\text{ when }}a>1;}&\scriptstyle {\log _{a}(+\infty )}&\scriptstyle {=}&\scriptstyle {+\infty .}\end{array}}$

The expressions in the second column are not to be considered as expressing numerical equalities ($\scriptstyle {\infty }$ not being a number); they are merely symbolical equations implying the relations indicated in the first column, and should be so understood.

22. Show that $\scriptstyle {{\underset {x=0}{\operatorname {limit} }}\;{\frac {\sin x}{x}}=1}$.^[1]

Let $\scriptstyle {O}$ be the center of a circle whose radius is unity.

Let $\scriptstyle {\operatorname {arc~} AM=\operatorname {arc~} AM'=x}$, and let $\scriptstyle {MT}$ and $\scriptstyle {M'T}$ be tangents drawn to the circle at $\scriptstyle {M}$ and $\scriptstyle {M'}$. From Geometry,

	$\scriptstyle {MPM'<MAM'<MTM'}$;
or	$\scriptstyle {2\sin x<2x<2\tan x}$.

Dividing through by $\scriptstyle {2\sin x}$, we get$\scriptstyle {1<{\frac {x}{\sin x}}<{\frac {1}{\cos x}}.}$

1. If we refer to the table on p. 4, it will be seen that for all angles less than 10° the angle in radians and the sine of the angle are equal to three decimal places. If larger tables are consulted, five-place, say, it will be seen that for all three angles less than 2.2° the sine of the angle and the angle itself are equal to four decimal places. From this we may well suspect that $\scriptscriptstyle {{\underset {x=0}{\operatorname {limit} }}\;{\frac {\sin x}{x}}=1.}$