University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises: 37

Answer

$\lim\limits_{x \to 1}\frac{x-1}{\sqrt {x+3} - 2} = 4$

Work Step by Step

$\frac{x-1}{\sqrt {x+3} - 2} = \frac{(x-1)(\sqrt {x+3} + 2)}{(\sqrt {x+3} - 2)(\sqrt {x+3} + 2)} = \frac{(x-1)(\sqrt {x+3} + 2)}{{x+3}-4} = \frac{(x-1)(\sqrt {x+3} + 2)}{(x-1)} = (\sqrt {x+3} + 2)$ Now, $\lim\limits_{x \to 1}\frac{x-1}{\sqrt {x+3} - 2} =\lim\limits_{x \to 1}(\sqrt {x+3} + 2) = 4$
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