\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=0.4in,right=0.5in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue} \large Lecture Notes} \chead{\color{black} \LARGE Trigonometric Limits} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ Hidegkuti, 2014} \rfoot{\small Last revised: June 27, 2015} \textwidth 7.6in \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*} \bigskip \bigskip \begin{center} {\LARGE Sample Problems\bigskip } \end{center} \begin{enumerate} \item Recall that $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=1$. \ Use this fact to compute each of the following limits. %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion a) \ $\lim\limits_{x\rightarrow 0}\dfrac{\sin 5x}{x}$ b) \ $\lim\limits_{x\rightarrow 0}\dfrac{1-\cos x}{x}$ c) \ $\lim\limits_{x\rightarrow 0}\dfrac{1-\cos x}{x^{2}}$ d) \ $\lim\limits_{x\rightarrow 0}\dfrac{\tan x}{x}$ e) \ $\lim\limits_{x\rightarrow 0}\dfrac{\sin 5x}{\sin 6x}$ f) \ $\lim\limits_{\theta \rightarrow \pi /4}\dfrac{\tan \theta -1}{~~\theta -\dfrac{\pi }{4}~~}$ g) \ $\lim\limits_{x\rightarrow 0}\dfrac{\sqrt{1-\cos x}}{x}$ \ %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \item a) \ Find the perimeter of a $15-$sided reguar polygon written into a circle with radius $10\unit{m}$. b) \ Find the perimeter of an $n-$sided regular polygon written into a circle with radius $R$. \ Use radians to measure angles. c) \ Find the limit of the perimeter of an $n-$sided regular polygon written into a circle with radius $R$ as $n$ approaches infinity. \ Use radians to measure angles. \item a) \ Find the area of a $15-$sided regular polygon written into a circle with radius $10\unit{m}$. b) \ Find the area of an $n-$sided regular polygon written into a circle with radius $R$. \ Use radians to measure angles. c) \ Find the limit of the area of an $n-$sided regular polygon written into a circle with radius $R$ as $n$ approaches infinity. \ Use radians to measure angles.{\LARGE \pagebreak } \end{enumerate} \begin{center} {\LARGE Practice Problems\bigskip } \end{center} \qquad Compute each of the following limits.% %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion \begin{enumerate} \item $\lim\limits_{x\rightarrow 0}\dfrac{\sin 5x}{x}$ \item $\lim\limits_{x\rightarrow 0}\dfrac{2\sin x\cos x}{x}$ \item $\lim\limits_{\theta \rightarrow 0}\dfrac{\cos 3\theta \sin 3\theta }{% \theta }$ \item $\lim\limits_{x\rightarrow 0}\dfrac{\sin 2x+\sin 4x}{x}$ \item $\lim\limits_{\theta \rightarrow 0}\dfrac{\tan 4\theta }{5\theta }$ \item $\lim\limits_{x\rightarrow 0}\dfrac{\sin 4x}{\sin 3x}$ \item $\lim\limits_{\theta \rightarrow 0}\dfrac{\theta }{\sin 3\theta }$ \item $\lim\limits_{x\rightarrow 0}\dfrac{\sin ^{2}x}{3x^{2}}$ \item $\lim\limits_{\theta \rightarrow 0}\dfrac{\sin \left( \theta ^{2}\right) }{\theta \tan \theta }$ \item $\lim\limits_{x\rightarrow 0}\dfrac{\tan 6x}{3x}$ \item $\lim\limits_{x\rightarrow 0}\dfrac{\sin x\tan x}{x^{2}}$ \item $\lim\limits_{x\rightarrow 0}\dfrac{\sin 2x\tan 3x}{x^{2}}$ \item $\lim\limits_{\theta \rightarrow 0}\dfrac{\theta \sin 2\theta }{% 2-2\cos ^{2}\theta }$ \item $\lim\limits_{x\rightarrow 0}\dfrac{x}{\sin 2x}$ \item $\lim\limits_{x\rightarrow 0}\dfrac{\tan x}{\tan 4x}$ \item $\lim\limits_{x\rightarrow 0}\dfrac{6x}{\sin 4x+\sin 3x}$ \item $\lim\limits_{\theta \rightarrow 0}\dfrac{2\theta ^{2}}{1-\cos \theta } $ \item $\lim\limits_{x\rightarrow \pi /2}\dfrac{\sin x-1}{x-\dfrac{\pi }{2}}$ \item $\lim\limits_{\theta \rightarrow 0}\dfrac{\theta \sin \theta }{1-\cos \theta }$ \end{enumerate} %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \bigskip \begin{center} {\LARGE Sample Problems - Answers\bigskip } \end{center} \begin{enumerate} \item a) \ $5\qquad $b) \ $0\qquad $c) \ $\dfrac{1}{2}\qquad $d) \ $1\qquad $% e) \ $\dfrac{5}{6}\qquad \ $f)$\ \ 2\qquad $g) \ $\func{undefined}$ \item a) \ $300\sin \left( \dfrac{\pi }{15}\right) \unit{m}\approx 62.\,\allowbreak 3735\unit{m}\qquad $b) \ $2nR\sin \dfrac{\pi }{n}\qquad $c) \ $2\pi R$ \item a) \ $750\sin \left( \dfrac{2\pi }{15}\right) \unit{m}^{2}\approx 305.\,\allowbreak 0525\unit{m}^{2}\qquad $b) \ $\dfrac{1}{2}nR^{2}\sin \dfrac{2\pi }{n}$ \qquad c) \ $\pi R^{2}${\LARGE \bigskip \bigskip } \end{enumerate} \begin{center} {\LARGE Answers - Practice Problems\bigskip \bigskip } \end{center} \qquad 1.) $\ 5\qquad $2.) $\ 2\qquad $3.) $\ 3\qquad $4.) $\ 6\qquad $5.) $% \ \dfrac{4}{5}\qquad $6.) $\ \dfrac{4}{3}\qquad $7.) $\ \dfrac{1}{3}\qquad $% 8.) $\ \dfrac{1}{3}\qquad $9.) $\ 1\qquad $10.) $\ 2\bigskip $ $\qquad $11.) $\ 1\qquad $12.) $\ 6\qquad $13.) $\ 1\qquad $14.) $\ \dfrac{1% }{2}\qquad $15.) $\ \dfrac{1}{4}\qquad $16.) $\ \dfrac{6}{7}\qquad $17.) $\ 4\qquad $18.) $\ 0\qquad $19.) $\ 2$ \bigskip \pagebreak \begin{center} {\LARGE Sample Problems - Solutions\bigskip } \end{center} \begin{enumerate} \item Recall that $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=1$. \ Use this fact to compute each of the following limits. a) \ $\lim\limits_{x\rightarrow 0}\dfrac{\sin 5x}{x}$% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{\sin 5x}{x}=\lim\limits_{x\rightarrow 0}% \dfrac{\sin 5x}{x}\cdot 1=\lim\limits_{x\rightarrow 0}\dfrac{\sin 5x}{x}% \cdot \dfrac{5}{5}=\lim\limits_{x\rightarrow 0}\dfrac{\sin 5x}{5x}\cdot 5=5\lim\limits_{x\rightarrow 0}\dfrac{\sin 5x}{5x} \end{equation*}% Let $y=5x.$ \ As $x$ approaches zero, so does $y$. \ So the limit becomes% \begin{equation*} 5\lim\limits_{x\rightarrow 0}\dfrac{\sin 5x}{5x}=5\lim\limits_{y\rightarrow 0}\dfrac{\sin y}{y}=5\cdot 1=5 \end{equation*}% b) \ $\lim\limits_{x\rightarrow 0}\dfrac{1-\cos x}{x}$% \begin{eqnarray*} \lim\limits_{x\rightarrow 0}\dfrac{1-\cos x}{x} &=&\lim\limits_{x\rightarrow 0}\dfrac{1-\cos x}{x}\cdot \dfrac{1+\cos x}{1+\cos x}=\lim\limits_{x% \rightarrow 0}\dfrac{1-\cos ^{2}x}{x\left( 1+\cos x\right) }% =\lim\limits_{x\rightarrow 0}\dfrac{\sin ^{2}x}{x\left( 1+\cos x\right) }% =\lim\limits_{x\rightarrow 0}\left( \dfrac{\sin x}{x}\cdot \dfrac{\sin x}{% 1+\cos x}\right) \\ &=&\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}\cdot \lim\limits_{x\rightarrow 0}\dfrac{\sin x}{1+\cos x}=1\cdot 0=0 \end{eqnarray*}% c) \ $\lim\limits_{x\rightarrow 0}\dfrac{1-\cos x}{x^{2}}$% \begin{eqnarray*} \lim\limits_{x\rightarrow 0}\dfrac{1-\cos x}{x^{2}} &=&\lim\limits_{x% \rightarrow 0}\dfrac{1-\cos x}{x^{2}}\cdot \dfrac{1+\cos x}{1+\cos x}% =\lim\limits_{x\rightarrow 0}\dfrac{1-\cos ^{2}x}{x^{2}\left( 1+\cos x\right) }=\lim\limits_{x\rightarrow 0}\dfrac{\sin ^{2}x}{x^{2}\left( 1+\cos x\right) }=\lim\limits_{x\rightarrow 0}\left( \dfrac{\sin ^{2}x}{x^{2}}\cdot \dfrac{1}{1+\cos x}\right) \\ &=&\lim\limits_{x\rightarrow 0}\dfrac{\sin ^{2}x}{x^{2}}\cdot \lim\limits_{x\rightarrow 0}\dfrac{1}{1+\cos x}=\left( \lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}\right) ^{2}\cdot \lim\limits_{x\rightarrow 0}\dfrac{1}{1+\cos x}=1^{2}\cdot \dfrac{1}{2}=% \dfrac{1}{2} \end{eqnarray*}% d) \ $\lim\limits_{x\rightarrow 0}\dfrac{\tan x}{x}$% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{\tan x}{x}=\lim\limits_{x\rightarrow 0}% \dfrac{~~\dfrac{\sin x}{\cos x}~~}{x}=\lim\limits_{x\rightarrow 0}\dfrac{% \sin x}{x\cos x}=\lim\limits_{x\rightarrow 0}\left( \dfrac{\sin x}{x}\cdot \dfrac{1}{\cos x}\right) =\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}\cdot \lim\limits_{x\rightarrow 0}\dfrac{1}{\cos x}=1\cdot 1=1 \end{equation*} e) \ $\lim\limits_{x\rightarrow 0}\dfrac{\sin 5x}{\sin 6x}$ Solution: \ We will bring this limit to a form where $\dfrac{\sin x}{x}$ appears.% \begin{eqnarray*} \lim\limits_{x\rightarrow 0}\dfrac{\sin 5x}{\sin 6x} &=&\lim\limits_{x% \rightarrow 0}\left( \dfrac{\sin 5x}{\sin 6x}\cdot \dfrac{x}{x}\right) =\lim\limits_{x\rightarrow 0}\left( \dfrac{\sin 5x}{x}\cdot \dfrac{x}{\sin 6x% }\right) =\lim\limits_{x\rightarrow 0}\dfrac{\sin 5x}{x}\cdot \lim\limits_{x\rightarrow 0}\dfrac{x}{\sin 6x} \\ &=&\lim\limits_{x\rightarrow 0}\left( \dfrac{\sin 5x}{x}\cdot \dfrac{5}{5}% \right) \cdot \lim\limits_{x\rightarrow 0}\left( \dfrac{x}{\sin 6x}\cdot \dfrac{6}{6}\right) =\lim\limits_{x\rightarrow 0}\left( \dfrac{\sin 5x}{5x}% \cdot 5\right) \cdot \lim\limits_{x\rightarrow 0}\left( \dfrac{6x}{\sin 6x}% \cdot \dfrac{1}{6}\right) \\ &=&5\lim\limits_{x\rightarrow 0}\dfrac{\sin 5x}{5x}\cdot \dfrac{1}{6}% \lim\limits_{x\rightarrow 0}\dfrac{6x}{\sin 6x}=5\cdot 1\cdot \dfrac{1}{6}% \cdot 1=\dfrac{5}{6} \end{eqnarray*}% \pagebreak f) \ $\lim\limits_{\theta \rightarrow \pi /4}\dfrac{\tan \theta -1}{~~\theta -\dfrac{\pi }{4}~~}$ This is clearly a $\dfrac{0}{0}$ type of an indeterminate. \ To simplify the denominator, we will introduce a new variable. \ Let $x=\theta -\dfrac{\pi }{% 4}$. \ As $\theta $ approaches $\dfrac{\pi }{4}$, $x$ will approach zero. \ Also, solving $x=\theta -\dfrac{\pi }{4}$ for $\theta $ we get $\theta =x+% \dfrac{\pi }{4}.$ So our limit becomes \begin{equation*} \lim\limits_{\theta \rightarrow \pi /4}\dfrac{\tan \theta -1}{~~\theta -% \dfrac{\pi }{4}~~}=\lim\limits_{x\rightarrow 0}\dfrac{\tan \left( x+\dfrac{% \pi }{4}\right) -1}{x} \end{equation*}% Now the denominator is simple, but the numerator became more complex. \ We will expand $\tan \left( x+\dfrac{\pi }{4}\right) $ using the sum formula for tangent.% \begin{eqnarray*} \lim\limits_{x\rightarrow 0}\dfrac{\tan \left( x+\dfrac{\pi }{4}\right) -1}{x% } &=&\lim\limits_{x\rightarrow 0}\dfrac{\dfrac{\tan x+\tan \dfrac{\pi }{4}}{% 1-\tan x\tan \dfrac{\pi }{4}}-1}{x}=\lim\limits_{x\rightarrow 0}\dfrac{% \dfrac{\tan x+1}{1-\tan x\cdot 1}-1}{x}=\lim\limits_{x\rightarrow 0}\dfrac{% \dfrac{\tan x+1}{1-\tan x}-\dfrac{1-\tan x}{1-\tan x}}{x} \\ &=&\lim\limits_{x\rightarrow 0}\dfrac{1}{x}\cdot \dfrac{\tan x+1-\left( 1-\tan x\right) }{1-\tan x}=\lim\limits_{x\rightarrow 0}\dfrac{1}{x}\cdot \dfrac{\tan x+1-1+\tan x}{1-\tan x}=\lim\limits_{x\rightarrow 0}\dfrac{1}{x}% \cdot \dfrac{2\tan x}{1-\tan x} \\ &=&\lim\limits_{x\rightarrow 0}\dfrac{\tan x}{x}\cdot \dfrac{2}{1-\tan x}% =\lim\limits_{x\rightarrow 0}\dfrac{\tan x}{x}\cdot \lim\limits_{x\rightarrow 0}\dfrac{2}{1-\tan x}=1\cdot 2=2 \end{eqnarray*}% g) \ $\lim\limits_{x\rightarrow 0}\dfrac{\sqrt{1-\cos x}}{x}$ This is also a $\dfrac{0}{0}$ type of an indeterminate. \ Solution 1: \ We will start by multiplying both numerator and denominator by $\sqrt{1+\cos x}$. \begin{eqnarray*} \lim\limits_{x\rightarrow 0}\dfrac{\sqrt{1-\cos x}}{x} &=&\lim\limits_{x% \rightarrow 0}\dfrac{\sqrt{1-\cos x}}{x}\cdot \dfrac{\sqrt{1+\cos x}}{\sqrt{% 1+\cos x}}=\lim\limits_{x\rightarrow 0}\dfrac{\sqrt{1-\cos x}\sqrt{1+\cos x}% }{x\sqrt{1+\cos x}}=\lim\limits_{x\rightarrow 0}\dfrac{\sqrt{\left( 1-\cos x\right) \left( 1+\cos x\right) }}{x\sqrt{1+\cos x}} \\ &=&\lim\limits_{x\rightarrow 0}\dfrac{\sqrt{1-\cos ^{2}x}}{x\sqrt{1+\cos x}}% =\lim\limits_{x\rightarrow 0}\dfrac{\sqrt{\sin ^{2}x}}{x\sqrt{1+\cos x}}% =\lim\limits_{x\rightarrow 0}\dfrac{\left\vert \sin x\right\vert }{x\sqrt{% 1+\cos x}} \end{eqnarray*}% If the expression was simply $\dfrac{\sin x}{x\sqrt{1+\cos x}}$, then we would be in a good shape, since $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x% }=1$. \ But we can not ignore that we have the absolute value of $\sin x$. \ We get rid of the absolute value sign by considering the sign of $\sin x.$ Case 1. \ If $x>0$, then also $\sin x>0$ and so $\left\vert \sin x\right\vert =\sin x$% \begin{equation*} \lim\limits_{x\rightarrow 0^{+}}\dfrac{\sqrt{1-\cos x}}{x}% =\lim\limits_{x\rightarrow 0^{+}}\dfrac{\left\vert \sin x\right\vert }{x% \sqrt{1+\cos x}}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin x}{x\sqrt{% 1+\cos x}}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin x}{x}\cdot \lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{\sqrt{1+\cos x}}=1\cdot \dfrac{1}{% \sqrt{2}}=\dfrac{1}{\sqrt{2}} \end{equation*}% Case 2. \ If $x<0$, then also $\sin x<0$ and so $\left\vert \sin x\right\vert =-\sin x$% \begin{equation*} \lim\limits_{x\rightarrow 0^{-}}\dfrac{\sqrt{1-\cos x}}{x}% =\lim\limits_{x\rightarrow 0^{-}}\dfrac{\left\vert \sin x\right\vert }{x% \sqrt{1+\cos x}}=\lim\limits_{x\rightarrow 0^{-}}\dfrac{-\sin x}{x\sqrt{% 1+\cos x}}=-1\cdot \lim\limits_{x\rightarrow 0^{-}}\dfrac{\sin x}{x}\cdot \lim\limits_{x\rightarrow 0^{-}}\dfrac{1}{\sqrt{1+\cos x}}=-1\cdot \dfrac{1}{% \sqrt{2}}=-\dfrac{1}{\sqrt{2}} \end{equation*}% Since the left-hand side limit and the right-hand side limit are different, the two-sided limit is undefined. Solution 2: \ Recall that $\sin ^{2}\dfrac{x}{2}=\dfrac{1-\cos x}{2}$ and so \ $1-\cos x=2\sin ^{2}\dfrac{x}{2}$.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{\sqrt{1-\cos x}}{x}=\lim\limits_{x% \rightarrow 0}\dfrac{\sqrt{2\sin ^{2}\dfrac{x}{2}}}{x}=\lim\limits_{x% \rightarrow 0}\dfrac{\left\vert \sqrt{2}\sin \dfrac{x}{2}\right\vert }{x} \end{equation*}% We will need to be a little bit careful because of the absolute value. \ If $% x$ is positive (recall it is also very close to zero) then so is $\sin \dfrac{x}{2}$. \ If $x$ is negative, so is $\sin \dfrac{x}{2}$. We will separately evaluate the left-side and right-side limits. \ \ Let us introduce the new variable $y=\dfrac{x}{2}$: \ \begin{eqnarray*} \lim\limits_{x\rightarrow 0^{+}}\dfrac{\sqrt{2}\left\vert \sin \dfrac{x}{2}% \right\vert }{x} &=&\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sqrt{2}\sin \dfrac{x}{2}}{x\cdot 1}=\sqrt{2}\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin \dfrac{x}{2}}{x\cdot 1}=\sqrt{2}\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin \dfrac{x}{2}}{x\cdot \dfrac{2}{2}}=\sqrt{2}\lim\limits_{x\rightarrow 0^{+}}% \dfrac{\sin \dfrac{x}{2}}{\dfrac{x}{2}\cdot 2}=\sqrt{2}\lim\limits_{x% \rightarrow 0^{+}}\dfrac{1}{2}\cdot \dfrac{\sin \dfrac{x}{2}}{\dfrac{x}{2}} \\ &=&\dfrac{\sqrt{2}}{2}\lim\limits_{y\rightarrow 0^{+}}\dfrac{\sin y}{y}=% \dfrac{\sqrt{2}}{2}\cdot 1=\dfrac{\sqrt{2}}{2} \end{eqnarray*}% The other side goes similarly:% \begin{eqnarray*} \lim\limits_{x\rightarrow 0^{-}}\dfrac{\sqrt{2}\left\vert \sin \dfrac{x}{2}% \right\vert }{x} &=&\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sqrt{2}\left( -\sin \dfrac{x}{2}\right) }{x\cdot 1}=-\sqrt{2}\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin \dfrac{x}{2}}{x\cdot 1}=-\sqrt{2}\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin \dfrac{x}{2}}{x\cdot \dfrac{2}{2}}=-\sqrt{2}% \lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin \dfrac{x}{2}}{\dfrac{x}{2}\cdot 2% } \\ &=&-\sqrt{2}\lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{2}\cdot \dfrac{\sin \dfrac{x}{2}}{\dfrac{x}{2}}=-\dfrac{\sqrt{2}}{2}\lim\limits_{y\rightarrow 0^{+}}\dfrac{\sin y}{y}=-\dfrac{\sqrt{2}}{2}\cdot 1=-\dfrac{\sqrt{2}}{2} \end{eqnarray*}% Since the right-hand side limit and the left-hand side limit are different, the two-sided limit is undefined. \item a) \ Find the perimeter of a $15-$sided reguar polygon written into a circle with radius $10\unit{m}$. \FRAME{itbpF}{1.67in}{1.5817in}{0in}{}{}{pic1.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.67in;height 1.5817in;depth 0in;original-width 0.1825in;original-height 0.1721in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'pic1.bmp';file-properties "XNPEU";}} \ \ \FRAME{itbpF}{1.6086in}{1.5645in}{0in}{}{}{pic1b.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.6086in;height 1.5645in;depth 0in;original-width 0.1583in;original-height 0.1531in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'pic1b.bmp';file-properties "XNPEU";}} \ \ \FRAME{itbpF}{1.2644in}{1.791in}{0in}{}{}{pic1c.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.2644in;height 1.791in;depth 0in;original-width 0.0493in;original-height 0.0813in;cropleft "0";croptop "1";cropright "1.1549";cropbottom "0";filename 'pic1c.bmp';file-properties "XNPEU";}} \ \FRAME{itbpF}{1.3206in}{2.1093in}{0in}{}{}{pic1d.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.3206in;height 2.1093in;depth 0in;original-width 0.0476in;original-height 0.0917in;cropleft "0";croptop "1";cropright "1.2032";cropbottom "0";filename 'pic1d.bmp';file-properties "XNPEU";}} The angle at the center of the circle is $\dfrac{360^{\circ }}{15}% =\allowbreak 24^{\circ }$ \ \ If we draw the altitude belonging to side $x$, we create a right triangle with an angle of $12^{\circ }$. \ From this right triangle, \ $\sin 12^{\circ }=\dfrac{~~~\dfrac{x}{2}~~~}{r}=\dfrac{x}{2r}$. \ So $x=2r\sin 12^{\circ }$. \ We convert the angle to radians and substitute $10\unit{m}$ for $r.$ \ The perimeter is the sum of all $15$ sides: \begin{equation*} P=15x=15\left( 2r\sin 12^{\circ }\right) =30\left( 10\unit{m}\right) \sin \left( \dfrac{\pi }{15}\right) =300\sin \left( \dfrac{\pi }{15}\right) \unit{% m}\approx 62.\,\allowbreak 373508\unit{m} \end{equation*}% \pagebreak b) \ Find the perimeter of an $n-$sided regular polygon written into a circle with radius $R$. \ Use radians to measure angles. Solution: \ We will perform the same steps as in the previous problem, only in the abstract. \ \ $x=2R\sin \left( \dfrac{2\pi }{2n}\right) =2R\sin \left( \dfrac{\pi }{n}\right) $. \ And so the perimeter of the polygon is% \begin{equation*} P=nx=n2R\sin \left( \dfrac{\pi }{n}\right) =2nR\sin \left( \dfrac{\pi }{n}% \right) \end{equation*}% c) \ Find the limit of the perimeter of an $n-$sided regular polygon written into a circle with radius $R$ as $n$ approaches infinity. \ Use radians to measure angles.% \begin{equation*} \lim\limits_{n\rightarrow \infty }\left( 2nR\sin \dfrac{\pi }{n}\right) =2R\lim\limits_{n\rightarrow \infty }\left( \dfrac{\sin \dfrac{\pi }{n}}{% \dfrac{1}{n}}\right) =2R\lim\limits_{n\rightarrow \infty }\left( \dfrac{\pi }{\pi }\cdot \dfrac{\sin \dfrac{\pi }{n}}{\dfrac{1}{n}}\right) =2\pi R\lim\limits_{n\rightarrow \infty }\left( \dfrac{\sin \dfrac{\pi }{n}}{% \dfrac{\pi }{n}}\right) \end{equation*}% Define $x=\dfrac{\pi }{n}$ . \ As $n\rightarrow \infty ,$ clearly $% x\rightarrow 0.$ \ Thus% \begin{equation*} 2\pi R\lim\limits_{n\rightarrow \infty }\left( \dfrac{\sin \dfrac{\pi }{n}}{% \dfrac{\pi }{n}}\right) =2\pi R\lim\limits_{x\rightarrow 0}\left( \dfrac{% \sin x}{x}\right) =2\pi R\cdot 1=2\pi R \end{equation*} \item a) \ Find the area of a $15-$sided regular polygon written into a circle with radius $10\unit{m}$. Solution: \ One can compute the altitude of the right triangle using right triangle trigonometry, and compute the area that way. \ However, we will use a moe efficient technique. \ Recall that the area of a triangle can be computed as $A=\dfrac{1}{2}ab\sin \gamma $ where $\gamma $ iis the angle between sides $a$ and $b.$ \ Then we can immediately compute the area of the isosceles triangle: \ $\dfrac{1}{2}R^{2}\sin 24^{\circ }$. \ So the area of the polygon is% \begin{equation*} A=15\left( \dfrac{1}{2}R^{2}\sin 24^{\circ }\right) =\dfrac{15}{2}\left( 10% \unit{m}\right) ^{2}\sin \left( \dfrac{2\pi }{15}\right) =750\sin \left( \dfrac{2\pi }{15}\right) \unit{m}^{2}\approx 305.\,\allowbreak 0525\unit{m}% ^{2} \end{equation*} b) \ Solution: \ We will perform the same steps as in the previous problem, only in the abstract. \ \begin{equation*} A=n\left( \dfrac{1}{2}R^{2}\sin \left( \dfrac{2\pi }{n}\right) \right) =% \dfrac{1}{2}nR^{2}\sin \left( \dfrac{2\pi }{n}\right) \end{equation*}% c) \ Find the limit of the area of an $n-$sided regular polygon written into a circle with radius $R$ as $n$ approaches infinity. \ Use radians to measure angles.% \begin{equation*} \lim\limits_{n\rightarrow \infty }\left( \dfrac{1}{2}nR^{2}\sin \dfrac{2\pi }{n}\right) =\lim\limits_{n\rightarrow \infty }\left( R^{2}\dfrac{\sin \dfrac{2\pi }{n}}{\dfrac{2}{n}}\right) =\lim\limits_{n\rightarrow \infty }\left( \pi R^{2}\dfrac{\sin \dfrac{2\pi }{n}}{\dfrac{2\pi }{n}}\right) =\pi R^{2}\lim\limits_{n\rightarrow \infty }\left( \dfrac{\sin \dfrac{2\pi }{n}}{% \dfrac{2\pi }{n}}\right) \end{equation*} \qquad Define $x=\dfrac{2\pi }{n}$ . \ As $n\rightarrow \infty ,$ clearly $% x\rightarrow 0.$ \ Thus% \begin{equation*} \pi R^{2}\lim\limits_{n\rightarrow \infty }\left( \dfrac{\sin \dfrac{2\pi }{n% }}{\dfrac{2\pi }{n}}\right) =\pi R^{2}\lim\limits_{x\rightarrow 0}\left( \dfrac{\sin x}{x}\right) =\pi R^{2}\cdot 1=\pi R^{2} \end{equation*} \end{enumerate} \bigskip \bigskip \bigskip {\small %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}