Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.2 Exercises - Page 887: 19

Answer

The limit is $$\lim_{(x,y)\to(0,1)}\frac{\arcsin xy}{1-xy}=0.$$ The function is discontinuous on a hyperbola $y=\frac{1}{x}$.

Work Step by Step

We will calculate this limit by substitution: $$\lim_{(x,y)\to(0,1)}\frac{\arcsin xy}{1-xy}=\frac{\arcsin(0\cdot1)}{1-0\cdot1}=\frac{\arcsin 0}{1}=0.$$ The function will have a discontinuity when $1-xy=0$ which is when $xy=1\Rightarrow y=1/x$. This is geometrically a hyperbola in the $xy$ plane. The discontinuity happens because when $1-xy=0$ the denominator iz $0$ and since we cannot divide by zero will have $$\lim_{(x,y)\to(x_0,y_0)}\frac{\arcsin xy}{1-xy}\neq\frac{\arcsin x_0y_0}{1-x_0y_0}.$$ At all other points from the domain we can just substitute and get $$\lim_{(x,y)\to(x_0,y_0)}\frac{\arcsin xy}{1-xy}=\frac{\arcsin x_0y_0}{1-x_0y_0},$$ so there the function would be continuous.
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