math solution 3

# math solution 3 - Homework due 7th February Solutions to...

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Homework due 7th February Solutions to compulsory questions Problem ( § 2.2 Exercise 8b). Compute the limit lim ( x,y ) (0 , 0) sin( xy ) y if it exists. Solution. Using the substitution u = xy we have lim ( x,y ) (0 , 0) sin( xy ) ( xy ) = lim u 0 sin u u = 1 . Therefore, by the properties of limits, lim ( x,y ) (0 , 0) sin( xy ) y = lim ( x,y ) (0 , 0) x × sin( xy ) ( xy ) = lim ( x,y ) (0 , 0) x lim ( x,y ) (0 , 0) sin( xy ) ( xy ) = 0 . Alternatively, by expanding sin( xy ) into a Taylor series we have lim ( x,y ) (0 , 0) sin( xy ) y = lim ( x,y ) (0 , 0) ( xy ) - 1 3! ( xy ) 3 + · · · y = lim ( x,y ) (0 , 0) x 1 - ( xy ) 2 3! + · · · = 0 . Problem ( § 2.2 Exercise 10c). Compute the limit lim ( x,y ) (0 , 0) ( x - y ) 2 x 2 + y 2 if it exists. Solution. Along the line y = x we have f ( x, x ) = ( x - x ) 2 x 2 + x 2 = 0 x 0 -→ 0 , and along the line y = - x we have f ( x, - x ) = ( x + x ) 2 x 2 + x 2 = 2 x 0 -→ 2 . Therefore the limit does not exist. Problem ( § 2.2 Exercise 15). If f : R n R and g : R n R are continuous, show that the function f 2 + g : R n -→ R x -→ [ f ( x )] 2 + g ( x ) is continuous. Solution. If a R n then lim x a ( f 2 + g )( x ) = lim x a ( [ f ( x )] 2 + g ( x ) ) (definition of f 2 + g ) = lim x a f ( x ) 2 + lim x a g ( x ) (using limit properties) = [ f ( a )] 2 + g ( a ) (since f, g continuous) = ( f 2 + g )( a ) (definition of f 2 + g ) . Therefore f 2 + g is continuous at each a R n , and so f 2 + g is continuous on R n . 1

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Problem ( § 2.3 Exercise 2ac). Evaluate the partial derivatives ∂z/∂x and ∂z/∂y for the functions: (a) z = a 2 - x 2 - y 2 at (0 , 0) and at ( a/ 2 , a/ 2). (c) z = e ax cos( bx + y ) at (2 π/b, 0). Solution. (a) We compute ∂z ∂x = - x a 2 - x 2 - y 2 and ∂z ∂y = - y a 2 - x 2 - y 2 Therefore ∂z ∂x (0 , 0) = ∂z ∂y (0 , 0) = 0 and ∂z ∂x ( a/ 2 , a/ 2) = ∂z ∂y ( a/ 2 , a/ 2) = - 1 2 .
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• Spring '08
• PARKINSON

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