15.2%20Ex%2026%20-%2030

15.2%20Ex%2026%20-%2030 - 628 C H A P T E R 15 D I F F E R...

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628 CHAPTER 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) 10. lim ( x , y ) ( 1 , 0 ) ln ( x y ) SOLUTION We use the continuity of ln ( x y ) at the point ( 1 , 0 ) to evaluate the limit by substitution: lim ( x , y ) ( 1 , 0 ) ln ( x y ) = ln ( 1 0 ) = ln 1 = 0 In Exercises 11–18, evaluate the limit or determine that it does not exist. 11. lim ( x , y ) ( 0 , 1 ) x y The function x y is continuous at the point ( 0 , 1 ) , hence we compute the limit by substitution: lim ( x , y ) ( 0 , 1 ) x y = 0 1 = 0 12. lim ( x , y ) ( 1 , 0 ) x y We examine the limit as ( x , y ) approaches ( 1 , 0 ) along the line x = 1: lim ( x , y ) ( 1 , 0 ) along x = 1 x y = lim y 0 1 y Because this limit does not exist, the limit lim ( x , y ) ( 1 , 0 ) x y does not exist. 13. lim ( x , y ) ( 1 , 1 ) e xy ln ( 1 + ) The function is continuous at ( 1 , 1 ) , hence we compute the limit using substitution: lim ( x , y ) ( 1 , 1 ) e ln ( 1 + ) = e 1 · 1 ln ( 1 + 1 · 1 ) = e ln 2 14. lim ( x , y ) ( 0 , 0 ) x 2 + y 2 1 + y 2 The function x 2 + y 2 1 + y 2 is continuous everywhere since it is a rational function whose denominator is never zero. We evaluate the limit using substitution: lim ( x , y ) ( 0 , 0 ) x 2 + y 2 1 + y 2 = 0 2 + 0 2 1 + 0 2 = 0 15. lim ( x , y ) ( 1 , 2 ) x 2 | y | 3 The function x 2 | y | 3 is continuous everywhere. We compute the limit using substitution: lim ( x , y ) ( 1 , 2 ) x 2 | y | 3 = ( 1 ) 2 |− 2 | 3 = 8 16. lim ( x , y ) ( 1 , 2 ) 2 | x | The function 2 | x | is the quotient of two continuous functions and the denominator is not zero at the point ( 1 , 2 ) . Therefore, the function is continuous at ( 1 , 2 ) and we may use substitution to compute the limit: lim ( x , y ) ( 1 , 2 ) 2 | x | = ( 1 )( 2 ) 2 |− 1 | =− 4 17. lim ( x , y ) ( 4 , 2 ) y 2 p x 2 4 The function is continuous at the point ( 4 , 2 ) , since it is the quotient of two continuous functions and the denominator is not zero at ( 4 , 2 ) . We compute the limit by substitution: lim ( x , y ) ( 4 , 2 ) y 2 p x 2 4 = 2 2 p 4 2 4 = 0 12 = 0
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SECTION 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 ( , 0 ) along the line x = and along the line y = x : lim ( x , y ) ( , 0 ) along x = sin x sin y = lim y 0 sin sin y = lim y 0 0 sin y = lim y 0 0 = 0 lim ( x , y ) ( , 0 ) along y = x sin x sin y = lim x sin x sin ( x ) = lim x sin x sin x = lim x ( 1 ) =− 1 The two limits are different, therefore the given limit does not exist.
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This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.

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15.2%20Ex%2026%20-%2030 - 628 C H A P T E R 15 D I F F E R...

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