\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.6in,right=0.7in,top=0.7in,bottom=0.7in} \pagestyle{fancy} \lhead{\color{blue} \large Lecture Notes} \chead{\color{black} \LARGE Two Important Trigonometric Limits} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ copyright Hidegkuti, Powell, 2012} \rfoot{\small Last revised: October 8, 2013} \textwidth 7.5in \textheight 9.6in \setlength{\headheight}{25pt} \setlength{\parindent}{0in} \begin{document} \fbox{Theorem 1: \ $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=1~$}% \bigskip Proof: \ This theorem and the next one are necessary for differentiating $% \sin x$ and $\cos x$. \ Recall a theorem: \ Let $\ r$ be the radius of a circle. \ If $\alpha $ is measured in radians, then the area of a sector with a central angle of $\alpha $ is $A_{\text{sector}}=\dfrac{\alpha r^{2}}{% 2}$. \ (Notation: \ $\overline{AB}$ will denote the length of line segment $% AB$.)$\medskip $ Let $x$ be a very small positive angle, measured in radians, drawn into a unit circle as shown on the picture below. \ Let $B$ be the point where the unit circle intersects the ray determined by $x$. \ We then draw a tangent line to the circle at point $B$. \ Let $A$ be the point where the tangent line intersects the $x-$axis. \ We also draw a vertical line through $B.$ \ Let $D$ be the point where this vertical line intersects the $x-$axis. \ Finally, let us denote by $E$ the point with coordinates $\left( 0,1\right) $% .\FRAME{dtbpF}{2.776in}{2.2399in}{0pt}{}{}{sinx_xbetter.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.776in;height 2.2399in;depth 0pt;original-width 2.7337in;original-height 2.2001in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'sinx_xbetter.bmp';file-properties "XNPEU";}} The proof will be based on the following fact: because they include each other, the following three areas can be easily compared: \begin{equation*} \text{Area of triangle }CDB\text{ }\leq \text{ Area of sector }CEB\text{ }% \leq \text{ Area of triangle }ABC \end{equation*} Area of triangle $CDB$: \ the horizontal side, $\overline{CD}=\cos x$ and the vertical side, $\overline{DB}=\sin x$. \ Since this is a right triangle, the area is: $A_{CDB}=\dfrac{1}{2}\sin x\cos x\medskip $ Area of sector $CEB$: $A_{\text{sector}}=\dfrac{1^{2}x}{2}=\dfrac{x}{2}% \medskip $ Area of triangle $ABC$: there is a right angle at point $B$ because the tangent line drawn to a circle is perpendicular to the radius drawn to the point of tangency. \ So the area is $A_{ABC}=\dfrac{1}{2}\overline{AB}\cdot \overline{BC}$. \ Clearly $\overline{BC}=1$. \ To compute $\overline{AB}$, in triangle $ABC$, \ $\tan x=\dfrac{\overline{AB}}{1}$ and so $\overline{AB}% =\tan x$.$\medskip $ Area of triangle $ABC$: \ $\dfrac{1}{2}\left( 1\right) \left( \tan x\right) =% \dfrac{\tan x}{2}$ or $\dfrac{\sin x}{2\cos x}$. \ So now \begin{equation*} \text{Area of triangle }CDB\text{ }\leq \text{ Area of sector }CEB\text{ }% \leq \text{ Area of triangle }ABC \end{equation*}% translates to \begin{equation*} \dfrac{1}{2}\sin x\cos x\leq \dfrac{x}{2}\leq \dfrac{\sin x}{2\cos x} \end{equation*} Let us divide all three sides by $\dfrac{\sin x}{2}$. \ Because $x$ is small and positive, $\dfrac{\sin x}{2}$ is positive and so we do not need to reverse the inequality signs.% \begin{equation*} \cos x\leq \dfrac{x}{\sin x}\leq \dfrac{1}{\cos x} \end{equation*}% Suppose now that $x$ approaches zero. \ Then both $\cos x$ and $\dfrac{1}{% \cos x}$ approach $1$. \ By the sandwich principle, $\dfrac{x}{\sin x}$, the quantity locked in between those two must also approach $1.$ \ \begin{equation*} \begin{array}{ccccc} \cos x & \leq & \dfrac{x}{\sin x} & \leq & \dfrac{1}{\cos x} \\ \downarrow & & & & \downarrow \\ 1 & & & & 1% \end{array}% \end{equation*}% If $\dfrac{x}{\sin x}$ approaches $1,$ so does its reciprocal, $\dfrac{\sin x% }{x}$.$\medskip $ So far, we have proven the statement for positive values of $x$, that is, $% \lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin x}{x}=1$. \ A similar argument works for negative values of $x$. \bigskip $\medskip \medskip $ \fbox{Theorem 2: \ $\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x}=0~$}% \bigskip Proof: \begin{eqnarray*} \lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x} &=&\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x}\cdot 1=\lim\limits_{x\rightarrow 0}\left( \dfrac{\cos x-1}{x}\cdot \dfrac{\cos x+1}{\cos x+1}\right) =\lim\limits_{x\rightarrow 0}% \dfrac{\cos ^{2}x-1}{x\left( \cos x+1\right) }=\lim\limits_{x\rightarrow 0}% \dfrac{-\left( 1-\cos ^{2}x\right) }{x\left( \cos x+1\right) } \\ &=&\lim\limits_{x\rightarrow 0}\dfrac{-\sin ^{2}x}{x\left( \cos x+1\right) }% =\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}\cdot \dfrac{-\sin x}{\cos x+1}% =\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}\cdot \lim\limits_{x\rightarrow 0}\dfrac{-\sin x}{\cos x+1}=1\cdot 0=0 \end{eqnarray*} $\vspace{4.25in}$ \bigskip \bigskip \href{http://www.teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html% }{For more documents like this, visit our page at\ http://www.teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}