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15.2Ex18-23

# 15.2Ex18-23 - S E C T I O N 15.2 Limits and Continuity in...

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S E C T I O N 15.2 Limits and Continuity in Several Variables (ET Section 14.2) 629 18. lim ( x , y ) ( π , 0 ) sin x sin y SOLUTION We examine the limits as ( x , y ) approaches ( π , 0 ) along the line x = π and along the line y = x π : lim ( x , y ) ( π , 0 ) along x = π sin x sin y = lim y 0 sin π sin y = lim y 0 0 sin y = lim y 0 0 = 0 lim ( x , y ) ( π , 0 ) along y = x π sin x sin y = lim x π sin x sin ( x π ) = lim x π sin x sin x = lim x π ( 1 ) = − 1 The two limits are different, therefore the given limit does not exist. In Exercises 19–22, assume that lim ( x , y ) ( 2 , 5 ) f ( x , y ) = 3 , lim ( x , y ) ( 2 , 5 ) g ( x , y ) = 7 19. lim ( x , y ) ( 2 , 5 ) ( f ( x , y ) + 4 g ( x , y ) ) SOLUTION Using the Sum Law and the Constant Multiples Law we get lim ( x , y ) ( 2 , 5 ) ( f ( x , y ) + 4 g ( x , y )) = lim ( x , y ) ( 2 , 5 ) f ( x , y ) + 4 lim ( x , y ) ( 2 , 5 ) g ( x , y ) = 3 + 4 · 7 = 31 20. lim ( x , y ) ( 2 , 5 ) f ( x , y ) g ( x , y ) 2 SOLUTION By the Product Law we have lim ( x , y ) ( 2 , 5 ) f ( x , y ) g ( x , y ) 2 = lim ( x , y ) ( 2 , 5 ) f ( x , y ) · lim ( x , y ) ( 2 , 5 ) g ( x , y ) 2 = 3 · 7 2 = 147 21. lim ( x , y ) ( 2 , 5 ) e f ( x , y ) SOLUTION e f ( x , y ) is the composition of G ( u ) = e u and u = f ( x , y ) . Since e u is continuous, we may evaluate the limit as follows: lim ( x , y ) ( 2 , 5 ) e f ( x , y ) = e lim ( x , y ) ( 2 , 5 ) f ( x , y ) = e 3 22. lim ( x , y ) ( 2 , 5 ) ln ( g ( x , y ) 2 f ( x , y ) ) SOLUTION The function is the composition G F , where G ( u ) = ln u and F ( x , y ) = g ( x , y ) 2 f ( x , y ) . We first use the Limit Laws to evaluate the limit: lim ( x , y ) ( 2 , 5 ) F ( x , y ) = lim ( x , y ) ( 2 , 5 ) ( g ( x , y ) 2 f ( x , y ) ) = lim ( x , y ) ( 2 , 5 ) g ( x , y ) 2 lim ( x , y ) ( 2 , 5 ) f ( x , y ) = 7 2 · 3 =
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