Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 36

Answer

$\dfrac{1}{2}$

Work Step by Step

Here, we have $\lim\limits_{y \to 0} f(0)=\dfrac{0}{0}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ Now, $p'(x)=(\dfrac{1}{2})(ay+a^2)^{-1/2}(a)=\dfrac{a}{2(ay+a^2)^{1/2}}$ and $q'(x)=1$ $ \lim\limits_{y \to 0} \dfrac{\dfrac{a}{2(ay+a^2)^{1/2}}}{1}=\dfrac{a}{2\sqrt {a(0)+a^2}}$ or, $\dfrac{a}{2a}=\dfrac{1}{2}$
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