University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Section 2.4 - One-Sided Limits - Exercises - Page 84: 12

Answer

$\lim_{x\to1^+}\sqrt\frac{x-1}{x+2}=0$

Work Step by Step

$$\lim_{x\to1^+}\sqrt\frac{x-1}{x+2}$$ To find one-side limit algebraically, we still apply the limit laws like for the two-side limits as normal. $$\lim_{x\to1^+}\sqrt\frac{x-1}{x+2}=\sqrt{\frac{1-1}{1+2}}=\sqrt\frac{0}{3}=\sqrt0=0$$ Therefore, $\lim_{x\to1^+}\sqrt\frac{x-1}{x+2}=0$
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