University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.5 - Indeterminate Forms and L'Hôpital's Rule - Exercises - Page 248: 35

Answer

$\dfrac{1}{2}$

Work Step by Step

L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$ Here, $\lim\limits_{y \to 0} f(0)=\dfrac{0}{0}$ This shows an indeterminate form of a limit, so we need to use L'Hospital's rule. Here, $A'(x)=\dfrac{1}{2}(5y+25)^{-1/2}(5)=\dfrac{5}{2(5+25)^{1/2}}$ and $B'(x)=1$ $ \lim\limits_{y \to 0} \dfrac{\dfrac{5}{2(5+25)^{1/2}}}{1}=\dfrac{5}{2\sqrt {5(0)+25}}=\dfrac{1}{2}$
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