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# hw17solutions - c On the other hand the subsequence...

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Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: March 19, 2010 Assignment 17 1. Let f : R R be deﬁned by setting f ( x ) := x if x is rational, and f ( x ) = 0 if x is irrational. (a) Show that f has a limit at x = 0. (b) Use a sequential argument to show that if c 6 = 0, then f does not have a limit at c . For part (a), the limit is 0. Indeed, for any given ε > 0, we set δ = ε . Then if 0 < | x | < δ , it follows that | f ( x ) | < ε = δ as either f ( x ) is 0 and this inequality holds trivially or f ( x ) = x and it follows from the assumption on x . For part (b), we can apply part (b) of Theorem 4.1.9. Indeed, for c 6 = 0, we construct a sequence x n c as follows. If n is even, we let x n by a rational number so that | x n - c | < 1 /n . If n is odd, we let x n be an irrational number with | x n - c | < 1 /n . The density of the rationals guarantees the existence of such an x n at every stage. We have that x n c easily. But then if we look at the subsequence of { f ( x n ) } consisting of even n , then f ( x n ) = x n
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Unformatted text preview: c . On the other hand, the subsequence consisting of the odd n has f ( x n ) = 0 → 0. Since c 6 = 0, we have that { f ( x n ) } does not converge as desired. 2. Let f,g be deﬁned on A ⊆ R to R , and let c be a cluster point of A . Suppose that f is bounded on a neighborhood of c and that lim x → c g = 0. Prove that lim x → c fg = 0. Fix ε > 0. Since f is bounded on a neighborhood of c , there exists a real number M and a δ 1 > 0 so that | f ( x ) | ≤ M if | x-c | < δ 1 and x ∈ A . Since lim x → c g = 0, we have that there exists a δ 2 > 0 so that | g ( x ) | < ε if 0 < | x-c | < δ 2 and x ∈ A . Setting δ = min( δ 1 ,δ 2 ), we see that | f ( x ) g ( x ) | ≤ M | g ( x ) | ≤ Mε if x ∈ A and 0 < | x-c | < δ . Thus, f ( x ) g ( x ) → 0. 1...
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