University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 13 - Section 13.2 - Limits and Continuity in Higher Dimensions - Exercises - Page 691: 21



Work Step by Step

Here, in this problem, the point $P(x,y) \to O(0,0)$ This means that $D(P,O) \to 0$ Therefore, $D^2=x^2+y^2; D \to 0$ Now, $\lim\limits_{D \to 0}\dfrac{\sin D^2}{D^2}=\dfrac{0}{0}$, which shows the limit of Indeterminate form; thus, we will apply L-Hospital's rule: We get $\lim\limits_{D \to 0}\dfrac{2 D\cos D^2}{2D}=\cos (0)=1$
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