Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises - Page 72: 66

Answer

The point $R(x_R,0)$ tends to $(-8,0)$ as $r$ tends to $0$.

Work Step by Step

The equation of circle $C_2$ is : $x^2+y^2 = r^2$ ...(1) And the equation of circle $C_1$ is : $(x-1)^2+y^2 = 1$ ... (2) To find the intersection point $Q$ we solve the quations (1) and (2) for $x$ and $y$. From (2) we have $x^2+y^2-2x+1 = 1$, or $x^2+y^2-2x = 0$. Now using (1) we get $r^2-2x =0$ or $x = \frac{r^2}{2}$. Substituting this value in (1) we get $y^{2} = r^2- \frac{r^4}{4} \Rightarrow y = \pm\sqrt{r^2- \frac{r^4}{4} }$. Since the $y$ coordinates of $Q$ is positive, we can write the coordinates of $Q$ as $(x,y) = ( \frac{r^2}{2},\sqrt{r^2- \frac{r^4}{4}})$. So the equation of line passing through $P(0,r)$ and $Q(\frac{r^2}{2},\sqrt{r^2- \frac{r^4}{4}})$ is $$\frac{y-r}{x-0} = \frac{\sqrt{r^2- \frac{r^4}{4}}-r}{\frac{r^2}{2}-0}.$$ To locate $R(x_R,0)$ we put $x = x_R, y =0$ in the equation of line $PQ$ to get, $$\frac{-r}{x_R} = \frac{\sqrt{r^2- \frac{r^4}{4}}-r}{\frac{r^2}{2}}$$ or $$x_R = -\frac{r^3}{2(\sqrt{r^2- \frac{r^4}{4}}-r)}.$$ Hence $$\lim\limits_{r \to 0}x_R = -\lim\limits_{r \to 0}\frac{r^3}{2(\sqrt{r^2- \frac{r^4}{4}}-r)}= -\lim\limits_{r \to 0}\frac{r^2}{2(\sqrt{1- \frac{r^2}{4}}-1)} ..(3)$$ Using binomial theorem for $|r|<1$we have, $\sqrt{1- \frac{r^2}{4}} = \Sigma_{k=0}^{\infty}\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k)}{k!}(- \frac{r^2}{4})^{k}$ $$\Rightarrow \sqrt{1- \frac{r^2}{4}} -1= \Sigma_{k=1}^{\infty}\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k)}{k!}(- \frac{r^2}{4})^{k}$$ $$\Rightarrow 2(\sqrt{1- \frac{r^2}{4}} -1)= 2\Sigma_{k=1}^{\infty}\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k)}{k!}(- \frac{r^2}{4})^{k}$$ $$\Rightarrow \frac{r^2}{2(\sqrt{1- \frac{r^2}{4}} -1)}=\frac{r^2}{ 2\Sigma_{k=1}^{\infty}\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k)}{k!}(- \frac{r^2}{4})^{k}}$$ $$=\frac{r^2}{ 2(\frac{1}{2})(\frac{1}{2}-1)(- \frac{r^2}{4})+2\Sigma_{k=2}^{\infty}\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k)}{k!}(- \frac{r^2}{4})^{k}}$$ $$=\frac{1}{ \frac{1}{8}+2\Sigma_{k=2}^{\infty}\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k)}{k!}(- \frac{r^2}{4})^{k-1}}$$ $$\Rightarrow \lim\limits_{r \to 0}\frac{r^2}{2(\sqrt{1- \frac{r^2}{4}}-1)}= \lim\limits_{r \to 0} \frac{1}{ \frac{1}{8}+2\Sigma_{k=2}^{\infty}\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k)}{k!}(- \frac{r^2}{4})^{k-1}} = \frac{1}{\frac{1}{8}}=8$$ Hence from (3) $$\lim\limits_{r \to 0}x_R = -\lim\limits_{r \to 0}\frac{r^2}{2(\sqrt{1- \frac{r^2}{4}}-1)} =-8$$ Hence the point $R(x_R,0)$ tends to $(-8,0)$ as $r$ tends to $0$.
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