## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Practice Exercises - Page 553: 77

#### Answer

$-2$

#### Work Step by Step

The Taylor series for $\cos x$ can be defined as: $\cos x=1-\dfrac{x^2}{2!}+\dfrac{ x^4}{4!}-....$ and the Taylor series for $\sin x$ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$ Now, $\lim\limits_{z \to 0} \dfrac{1-\cos^2 z}{\ln (1-z) +\sin z}=\lim\limits_{z \to 0} \dfrac{1- (1-z^2+z^4/3-...)}{(-z-\dfrac{z^2}{2}-....)+(z-\dfrac{z^3}{3!}+.....}$ or, $=\lim\limits_{z \to 0} \dfrac{1-\dfrac{z^2}{3}+...}{-\dfrac{1}{2}-\dfrac{z}{3}-....}$ or, $=\dfrac{1}{-1/2}$ or, $=-2$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.