there exist a neighborhood U x of x and a number M x on U x Now U x x is an

There exist a neighborhood u x of x and a number m x

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, there exist a neighborhood U x of x and a number M x on U x . Now {U x : x ∈ 𝐷} is an open cover for D since D is compact, and there Exist X 1, x 2 , xn in D such that D U x1 U xn. . Let M = max{ Mx 1 , ..Mx n } then f is bounded by M on D. 15b) Let f(x) = 1/x and D =(0, 1) F is continuous on D and bounded on each neighborhood of each x in D, but D is not compact here. Here f is not bounded on D because D is open and not bounded.
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Additional Problems: 1) Yes f is continuous when A = -3 and when A =1 because this is where the limit from the left, right and at c = 1, are equivalent. 2) For f(x) to be continuous at x = 1, then limit from the left and the right must be equivalent. Therefore, A=3 and B = 0. If the functions are dysfunctional at x=2, then B must not be ¾ when A= -3
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  • Fall '08
  • Staff
  • Calculus, Continuous function, Limit of a function, continuous ∀D465

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