University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Additional and Advanced Exercises - Page 114: 28



Work Step by Step

$$A=\lim_{x\to0}\frac{\sin(x^2+x)}{x}$$ $$A=\lim_{x\to0}\frac{\sin(x^2+x)}{x^2+x}\times\lim_{x\to0}\frac{x^2+x}{x}$$ For $x\ne0$ and $x\ne-1$, $x^2+x\ne0$. So, $\lim_{x\to0}\frac{\sin(x^2+x)}{x^2+x}=1$. $$A=1\times\lim_{x\to0}\frac{x^2+x}{x}$$ $$A=\lim_{x\to0}(x+1)$$ $$A=0+1=1$$
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