Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.2 - Limits and Continuity - 14.2 Exercise - Page 911: 41

Answer

-1

Work Step by Step

Conversion of polar co-ordinates $(r, \theta)$: $x=r \cos \theta$ and $y= r \sin \theta$ Given: $\lim\limits_{(x,y) \to(0,0)}\dfrac{e^{-x^2-y^2}-1}{(x^2+y^2)}$ This implies that $=\lim\limits_{r \to0}\dfrac{e^{(-r^2 \cos^2 \theta-r^2 \sin^2 \theta)}-1}{(r^2 \cos^2 \theta+r^2 \sin^2 \theta)}$ $=\lim\limits_{r \to0}\dfrac{e^{(-r^2 \cos^2 \theta-r^2 \sin^2 \theta)}-1}{r^2}$ Using L-Hospital's Rule. $:=\lim\limits_{r \to 0}\dfrac{-2re^{-r^2}}{2r}$ $=-1$
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