assignment2_sol

# assignment2_sol - MATH1211/10-11(2)/Asol2/TNK Department of...

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Unformatted text preview: MATH1211/10-11(2)/Asol2/TNK Department of Mathematics, The University of Hong Kong MATH1211 (2010–11, 2nd semester) Suggested solution to Assignment 2 Note : Some solutions given below are outline only. You may need to give more details in your solution. The parts that are typed in red color are remarks for you to read; they are not a part of the solution. 1. (a) Evaluating the limit along the line x = 0, we obtain lim ( x,y ) → (0 , 0) ,x =0 2 xy x 2 + y 2 = lim y → y 2 = 0 . Evaluating the lime along the line x = y , we obtain lim ( x,y ) → (0 , 0) ,x = y 2 xy x 2 + y 2 = lim x → 2 x 2 x 2 + x 2 = 1. Because these two limits are not equal, we conclude that lim ( x,y ) → (0 , 0) ,x =0 2 xy x 2 + y 2 does not exist. (Remark : You may use polar form to solve this problem.) (b) Note that when ( x,y,z ) → (0 , , 0), we have x → 0, y → 0, and z → 0 respectively. Note also that when ( x,y,z ) 6 = (0 , , 0), ≤ y 3 x 2 + y 2 + z 4 = | y | y 2 x 2 + y 2 + z 4 ≤ | y | . Here the last inequality follows because (0 ≤ ) y 2 x 2 + y 2 + z 4 ≤ 1. Thus when ( x,y,z ) → (0 , , 0), we have y → 0, and hence y 3 x 2 + y 2 + z 4 → 0 also, i.e., lim ( x,y,z ) → (0 , , 0) y 3 x 2 + y 2 + z 4 = 0. Similarly, we have ≤- 1000 xy 2 x 2 + y 2 + z 4 = 1000 | x | y 2 x 2 + y 2 + z 4 ≤ 1000 | x | , ≤ z 5 x 2 + y 2 + z 4 = | z | z 4 x 2 + y 2 + z 4 ≤ | z | , and hence lim ( x,y,z ) → (0 , , 0)- 1000 xy 2 x 2 + y 2 + z 4 = lim ( x,y,z ) → (0 , , 0) z 5 x 2 + y 2 + z 4 = 0. By adding these three limits (see Theorem 2.5) we have lim ( x,y,z ) → (0 , , 0) y 3- 1000 xy 2 + z 5 x 2 + y 2 + z 4 = 0 + 0 + 0 = 0 . 2. We first examine the value of f ( x,y ) when ( x,y ) → (0 , 0) along straight lines with equation y = mx where m is any real number. In fact lim ( x,y ) → (0 , 0) ,y = mx f ( x,y ) = lim x → x 4 m 4 x 4 ( x 2 + m 4 x 4 ) 3 = lim x...
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## assignment2_sol - MATH1211/10-11(2)/Asol2/TNK Department of...

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