Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises: 36

Answer

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

Work Step by Step

$$A=\lim_{x\to1}\frac{x\sin(x-1)}{2x^2-x-1}$$ As $x\to1$, $[x\sin(x-1)]\to[1\times\sin(1-1)]=1\times\sin0=0$ and $(2x^2-x-1)\to(2\times1^2-1-1)=2-2=0.$ Therefore, we have here an indeterminate form of $\frac{0}{0}$. L'Hospital's Rule would thus be applied: $$A=\lim_{x\to1}\frac{[x\sin(x-1)]'}{[2x^2-x-1]'}$$ $$A=\lim_{x\to1}\frac{\sin(x-1)+x\cos(x-1)}{4x-1}$$ $$A=\frac{\sin(1-1)+1\times\cos(1-1)}{4\times1-1}$$ $$A=\frac{\sin0+1\times\cos0}{3}$$ $$A=\frac{0+1\times1}{3}$$ $$A=\frac{1}{3}$$
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