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: 29



Work Step by Step

$$A=\lim_{x\to2}\frac{\sin(x^2-4)}{x-2}$$ $$A=\lim_{x\to2}\frac{\sin(x^2-4)}{x^2-4}\times\lim_{x\to2}\frac{x^2-4}{x-2}$$ For $x\ne\pm2$, $x^2-4\ne0$ and $\lim_{x\to2}(x^2-4)=0$. So we have $\lim_{x\to2}\frac{\sin(x^2-4)}{x^2-4}=1$. $$A=1\times\lim_{x\to2}\frac{(x-2)(x+2)}{x-2}$$ $$A=\lim_{x\to2}(x+2)$$ $$A=2+2=4$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.