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Tutorial2_sol

# Tutorial2_sol - MATH1211/10-11(1/Tu2sol Department of...

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MATH1211/10-11(1))/Tu2sol Department of Mathematics, The University of Hong Kong MATH1211 Multivariable Calculus (2010–2011, 1st semester) Solution to Tutorial 2 1. (a) Since lim ( x,y ) ( - 1 , 2) 2 x 2 + y 2 = 2 + 4 = 6 and lim ( x,y ) ( - 1 , 2) x 2 + y 2 = 1 + 4 = 5, it follows that lim ( x,y ) ( - 1 , 2) 2 x 2 + y 2 x 2 + y 2 = 6 5 . (b) (Method 1) When considering the limit along the line x = 0: lim ( x,y ) (0 , 0) ,x =0 x 2 x 2 + y 2 = lim y 0 0 0 + y 2 = 0 . When considering the limit along the line y = 0: lim ( x,y ) (0 , 0) ,y =0 x 2 x 2 + y 2 = lim x 0 x 2 x 2 = 1 . Since the function has different limiting values when ( x, y ) (0 , 0) along different paths, we conclude that the limit actually does not exist. (Method 2) Let x = r cos θ, y = r sin θ . Then lim ( x,y ) (0 , 0) x 2 x 2 + y 2 = lim r 0 + r 2 cos 2 θ r 2 = cos 2 θ, which has different values depending on θ . Hence the limit does not exist. 2. We first find f ( x, y ) and then plug in ( x, y ) = (2 , 1): f ( x, y ) = ( f x , f y ) = ye xy + 1 x - y , xe xy - 1 x - y , f (2 , 1) = ( e 2 + 1 , 2 e 2 - 1) .

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Tutorial2_sol - MATH1211/10-11(1/Tu2sol Department of...

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