assignment2_sol

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online