Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises: 50

Answer

$$\lim\limits_{x\to1}\frac{\sin(x-1)}{x^2+x-2}=\frac{1}{3}$$

Work Step by Step

$$A=\lim\limits_{x\to1}\frac{\sin(x-1)}{x^2+x-2}$$$$A=\lim\limits_{x\to1}\frac{\sin(x-1)}{(x-1)(x+2)}$$ $$A=\lim\limits_{x\to1}\frac{\sin(x-1)}{x-1}\times\lim\limits_{x\to1}\frac{1}{x+2}$$ Let $x-1=\theta$. Since $x\to 1$ is considered, $(x-1)\to0$, so $\theta\to0$ also. $$A=\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta}\times\lim\limits_{x\to1}\frac{1}{x+2}$$ $$A=1\times\frac{1}{1+2}=\frac{1}{3}$$
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