1873_solutions

# X 1 and x 2 such that 3 x 1 3 x 2 3 we observe

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Unformatted text preview: d observe that, whenever 0, 0   we have |f x, y |  . x, y 0, 0 and x, y b. For each point x, y 0, 0 we define f x, y  x2y .  y2 x4 0.5 0 -0.5 4 2 y 0 x 0 2 4 Whenever y 0 we have f 0, y  0 and so, if f has a limit at 0, 0 then this limit is 0. Note also that if x 0 then f x, 0  0 which again suggests that, if there is a limit, then the limit is 0. However, whenever x 0 we have x2x2 f x, x 2  1 4  x2 2 2 x and so,if there is a limit, then the limit must be 1 . Therefore f has no limit at the point 0, 0 . 2 The figure illustrates how the function values remain at 1 as x, y 0, 0 along the parabola 2 y  x2. c. For each point x, y 0, 0 we define f x, y  211 x4y4 .  y6 x6 12 10 8 6 4 2 0 -5 y 0 5 If |x | -4 0 x 2 4 |y |, then y8 x4y4  y2 6 y y6 and if |y | |x |, then f x, y x 2 . Now we can show that f x, y 0 as x, y 0, 0 . Suppose that  0 and define  to be the smaller of the two numbers 1 and . Then, whenever x, y 0, 0 and x, y   we have . |f x, y |   2 f x, y  d. For each point x, y x6 0, 0 we define f x, y  xy x 2 y 2 . x2  y2 6 4 2 0 -2 -4 -6 0 y 5 4 2 -4 -2 x Since xy x2  y2 whenever x, y 1 2 0, 0 we have x2  y2 . 2 2 Now we can show that f x, y 0 as x, y 0, 0 . Suppose that  0. We define  to be the smaller of the numbers 1 and . Whenever x, y 0, 0 and x, y  we have |f x, y |  . |f x, y | e. For each point x, y |x 2 y2 | 0, 0 we define f x, y  212 xy x 2 x2  y2 y2 3/2 . 1 0.5 0 -0.5 -1 -2 y 0 2 4 4 -4 -2 x 0 2 Since xy x2  y2 whenever x, y 0, 0 we have Now we can show that f x, y Whenever x, y 0, 0 and x2  y2 2 2 y2 | 1 |x 2 x2  y2 |f x, y | f. For each point x, y 1 2 0 as x, y 0, 0 . Suppose that x, y  we have |f x, y |  .  0. We define   . 0, 0 we define xy x 2 f x, y  y2 x2  y2 2 . 0.2 0.1 0 -0.1 -0.2 0 y -4 4 4 x 2 Whenever x 0 we have f x, 0  0. Therefore, if f has a limit at 0, 0 then the limit is zero. However, if x 0 then f x, x 2  x x 2 x2 x2  x2 4 Therefore if f has a limit at 0, 0 then the limit must be 0, 0 . g. For each point x, y x2 4 2 6 25  6. 25 and we conclude that f has no limit at 0, 0 we define f x, y  213 x2y2 x2 x2  y2 y2 2 . 3 2 1 0 -1 -2 -3 -5 y 5 Whenever x, y -4 -2 2 4 x 0, 0 , since x2y2 x2  y2 1 4 2 we have |f x, y | and, arguing as above, we see that f x, y h. For each point x, y 1 |x 2 y 2 | 4 0 as x, y 0, 0 . 0, 0 we define f x, y  xy 2 x 2 y2 . 2 x2  y2 1 0 -1 -4 -4 -2 -2 y 0 x 0 2 2 4 Whenever x, y 4 0, 0 we have |f x, y |   xy 2 x 2 y2 x2  y2 2 xy 2 x  y x y 2 x2  y2 x2y2 x y 2 x2  y2  xy 3 x y 2 x2  y2 and we can argue as in the preceding exercises to show that each of these terms approaches 0 as x, y 0, 0 . i. For each point x, y 0, 0 we define f x, y  214 x2y2 3/4 x2  x2 y2 2 y2 . 0.5 0 -0.5 0 y 5 Whenever x, y 4 x -4 -2 0, 0 we have |x | 3/2 |y | 3/2 |x |  |y | |x |  |y | |f x, y | 2 x2  y2 |x | 3/2 |y | 3/2 |x | |x |  |y | |x | 3/2 |y | 3/2 |y | |x |  |y |  2 2 x2  y2 x2  y2 |x | 3/2 |y | 3/2 |x ||y | |x | 3/2 |y | 3/2 |y ||x | |x | 5/2 |y | 5...
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