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  xc  < 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 <  xc  < 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 <  xc  < δ . Thus, f ( x ) g ( x ) → 0. 1...
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 Spring '10
 MetCalfe
 Calculus, Irrational number, Xn, sequence Xn, J. Metcalfe

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