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Unformatted text preview: 1 Using the limit deﬁnition to show a limit exists.
xy 2 . We want to show x2 + y 2 xy 2 = 0. (x,y )→(0,0) x2 + y 2 lim Let f (x, y ) = The deﬁnition says that we need to show |f (x, y ) − L| = like, when (x, y ) is suﬃciently close to the origin. xy 2 − 0 gets as small as we x2 + y 2 To do this, we will estimate an upper bound for this quantity. √ Step one: notice |x| = x2 . √ Step one and a half: notice |x| = x2 ≤ x2 + y 2 . Step two: notice y 2 ≤ x2 + y 2 . Putting all this together, we have |f (x, y ) − L| = xy 2 |x|y 2 =2 ≤ x2 + y 2 x + y2 x2 + y 2 (x2 + y 2 ) = x2 + y 2 x2 + y 2 . So the quantity |f (x, y ) − L| is at most the distance of the point (x, y ) from the origin. This means that as long as (x, y ) is a point inside the disk of radius centered at (0, 0), then |f (x, y ) − L| < .
1 For example, if we look at any point (x, y ) inside the disk of radius 10 , then the diﬀerence 1 1 1 between f (x, y ) and the limit L = 0 is at most 10 . (Try computing f ( 20 , 20 ) and see that it 1 is less than 10 .) If we look at any point (x, y ) inside the disk of radius 1 f (x, y ) and the limit L = 0 is at most 2000 . Therefore, the limit of f (x, y ) is 0. 1 2000 , then the diﬀerence between ...
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This note was uploaded on 03/28/2011 for the course MATH 20C taught by Professor Helton during the Winter '08 term at UCSD.
- Winter '08