Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - Problems Plus - Problems - Page 827: 4


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

Work Step by Step

We have $\lim\limits_{n \to \infty}(\frac{2^{n-1}}{\sqrt {4^n+\frac{2}{3}}})=\sqrt {\lim\limits_{n \to \infty}(\frac{3.2^{(-2)}}{1+2^{1-2n}}})=\frac{\sqrt 3}{2}$ Hence, $$\lim\limits_{n \to \infty}\angle P_nAP_n+1=\frac{\pi}{3}$$
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