ma100_hw5

Ma100_hw5 - = lim r-→ cos θ sin θ -r 2 = 0 Since cos θ and sin θ are bounded functions we can take limit as r-→ directly So the limit of

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Homework answer 5 Sec 14.2 26(a). Along the curve y = mx , lim ( x,y ) -→ (0 , 0) x 3 y 2 x 6 + y 2 = lim x -→ 0 x 3 ( mx ) 2 x 6 + ( mx ) 2 = lim x -→ 0 mx 4 2 x 6 + m 2 x 2 = 0 , m = 0; lim x -→ 0 mx 2 2 x 4 + m 2 , m 6 = 0 . = 0 Along the curve y = kx 2 where k 6 = 0, lim ( x,y ) -→ (0 , 0) x 3 y 2 x 6 + y 2 = lim x -→ 0 x 3 ( kx 2 ) 2 x 6 + ( kx 2 ) 2 = lim x -→ 0 kx 5 2 x 6 + k 2 x 4 = lim x -→ 0 kx 2 x 2 + k 2 = 0 So it tends to 0 along the curve y = mx or y = kx 2 . (b). Along the curve y = x 3 , lim ( x,y ) -→ (0 , 0) x 3 y 2 x 6 + y 2 = lim x -→ 0 x 3 ( x 3 ) 2 x 6 + ( x 3 ) 2 = lim x -→ 0 x 6 3 x 6 = 1 3 So the limit does not exist as ( x, y ) -→ (0 , 0) since we got different limits along different curves. 1
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2 32. Substitute x = r cos θ and y = r sin θ , lim ( x,y ) -→ (0 , 0) xy ln( x 2 + y 2 ) = lim r -→ 0 + r 2 cos θ sin θ ln r 2 = lim r -→ 0 + cos θ sin θ · 2 r r 2 - 2 r 3 , (by L’Hopital’s Rule)
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Unformatted text preview: = lim r-→ + cos θ sin θ · (-r 2 ) = 0 Since cos θ and sin θ are bounded functions, we can take limit as r-→ + directly. So the limit of function f ( x, y ) at (0 , 0) exists. Then, we can define the function f ( x, y ) = ± xy ln( x 2 + y 2 ) , for ( x, y ) 6 = (0 , 0); , for ( x, y ) = (0 , 0). so that f ( x, y ) will be continuous....
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This note was uploaded on 09/30/2010 for the course MATH 100 taught by Professor Qt during the Fall '09 term at HKUST.

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Ma100_hw5 - = lim r-→ cos θ sin θ -r 2 = 0 Since cos θ and sin θ are bounded functions we can take limit as r-→ directly So the limit of

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