1873_solutions

X is close to 2 we shall use this observation to

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Unformatted text preview: V is a neighborhood of . Using condition c we choose a number   0 such that the condition f x V will hold whenever x S a , a   a . Since the interval a , a   is a neighborhood of a, condition a must hold. 7. Given that S is a subset of a metric space X, that f is a function from X into a metric space Y, that  that a is an interior point of S, prove that the following conditions are equivalent: a. f x  as x Y and a. b. For every number  0 there exists a number   0 such that the inequality d f x ,   every point x X a that satisfies the inequality d x, a  . 209 will hold for The only way in which condition b differs from the ,  form of the assertion that f x  as x a is that it requires that d f x ,   for all points x within a distance  of a and unequal to a. It does not merely assert that d f x ,   when x is a member of the set S unequal to a and within a distance  of a. It is obvious that condition b implies condition a. To show that condition a implies condition b we assume that f x  as x a. Suppose that  0. Choose a number  1  0 such that the condition d f x ,   will hold whenever x S B a,  1 a . Now, using the fact that a is an interior point of S, choose a number  2  0 such that B a,  2 S. We define  to be the smaller of the two numbers  1 and  2 and we observe that the inequality d f x ,   will hold for every point x that satisfies the inequality d a, x  . 8. Suppose that S is a set of real numbers, that a is a limit point of S, that f : S R and that  is a real number. Complete the following sentence: The function f fails to have a limit of  at the number a when there exists a neighborhood V of  such that ...... The function f fails to have a limit of  at the number a when there exists a number  0 such that for every number   0 there is at least one number x in the set S a , a   a for which . |f x  | 9. Prove that 1 x  xy 1  x 2  xy  y 2 as x, y 1, 2 . We begin with the observation that if u  x 1 x  xy 1  x 2  xy  y 2 1 4  1 4 1 and v  y 2 then 1 u1  u1 v2 1 u1 2 u1 v2  v2 2 1 4 u 2  v 2  v 3uv 1 . 4 8  u 2  4u  uv  5v  v 2 In the event that |x 2 |  1 and |y 2 |  1 we see that 2 2 1 1 2  4u  uv  5v  v 2 51 3 |8  u | 8  u2  v2 4 1 2 2 2 2 and so |u 2  v 2  v 3uv | u 2  v 2  v 3uv 1 . 4 8  u 2  4u  uv  5v  v 2 12 Suppose that  0 and define  to be the smaller of the two numbers 1 and . Then whenever 2 1, 2  , defining, u  x 1 and v  y 2, we have x, y 1, 2 and x, y |u 2  v 2  v 3uv | 1 x  xy 1 6 . 2  xy  y 2 4 12 12 1x  10. In each of the following cases, determine whether or not the function f has a limit at 0, 0 . Use Scientific Notebook to view the graph of the function f in each case. a. For each point x, y 0, 0 we define f x, y  210 x2y .  y2 x2 2 0 -2 y 2 -2 x 4 -4 4 Since x2 1 x2  y2 whenever x, y 0, 0 we have |f x, y | |y | whenever x, y 0, 0 . Now we can show that f x, y 0 as x, y 0, 0 . Suppose that  0. We define   an...
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