University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Practice Exercises - Page 111: 30

Answer

$$\lim_{x\to\sqrt5}g(x)=\frac{1}{2}-\sqrt5$$

Work Step by Step

$$\lim_{x\to\sqrt5}\frac{1}{x+g(x)}=2$$ $$\frac{1}{\lim_{x\to\sqrt5}(x+g(x))}=2$$ $$\lim_{x\to\sqrt5}(x+g(x))=\frac{1}{2}$$ $$\sqrt5+\lim_{x\to\sqrt5}g(x)=\frac{1}{2}$$ $$\lim_{x\to\sqrt5}g(x)=\frac{1}{2}-\sqrt5$$
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