Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Problems Plus - Problems: 4


$\lim\limits_{n \to \infty}\angle P_nAP_{n+1}=\frac{\pi}{3}$

Work Step by Step

$\lim\limits_{n \to \infty}\frac{2^{n-1}}{\sqrt {4^n+2}/3}$ $=\sqrt {\lim\limits_{n \to \infty}\frac{3.2^{(-2)}}{1+2^{1-2n}}}$ $=\frac{\sqrt 3}{2}$ Thus, $\lim\limits_{n \to \infty}\angle P_nAP_{n+1}=\frac{\pi}{3}$
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