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 - Page 67: 36


$\lim\limits_{x \to 4}\frac{4x-x^{2}}{2-\sqrt x} = 16$

Work Step by Step

$\frac{4x-x^{2}}{2-\sqrt x} = \frac{x(4-x)}{2-\sqrt x} = \frac{x(2-\sqrt x)(2+\sqrt x)}{2-\sqrt x} = x(2+\sqrt x)$ Now, $\lim\limits_{x \to 4}\frac{4x-x^{2}}{2-\sqrt x} = \lim\limits_{x \to 4}x(2+\sqrt x) = 4(2+2)= 16$
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