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Unformatted text preview: MATH 413 HW 5: Solution to selected problems. October 2, 2007 4.1.5.4 Give a definition of lim x → + ∞ f ( x ) = y . Show that this is true if and only if for every sequence x 1 ,x 2 ,... of points in the domain of f such that lim n → + ∞ x n = + ∞ we have lim n → + ∞ f ( x n ) = y . Solution: Suppose that f is a function with domain D unbounded from above, we say that lim x → + ∞ f ( x ) = y , where y ∈ R , if for all > 0 there exists M ∈ R such that ( x ∈ D ∧ x > M ) ⇒ (  f ( x ) y  < ). Observe that if f has a domain bounded from above then lim x → + ∞ f ( x ) = y for any y ∈ R , a situation that we would like to avoid. Below we will assume that the domain of f is unbounded from above. (i) Assume that lim x → + ∞ f ( x ) = y holds according to our definition, and let n 7→ x n be an arbitrary sequence in D with lim n → + ∞ x n = ∞ . Given > 0, by definition there exists M > 0 such that ( x ∈ D ∧ x > M ) ⇒ (  f ( x ) y  < ). Since the sequence n 7→ x n diverges to infin ity as n goes to infinity, we conclude that there exists N ∈ N such that ( n > N ) ⇒ ( x n > M ). By combining the statements above we get ( n > N ) ⇒ (  f ( x n ) y  < ). Then lim n → + ∞ f ( x n ) = y . (ii) Now assume that lim x → + ∞ f ( x ) = y does not hold. This means that there exists > 0 such that for all M ∈ R there is x ∈ D such that x > M and  f ( x ) y  > . This assures the existence of a sequence n 7→ x n such that x n ∈ D , x n > n and  f ( x n ) y  > for all n ∈ N . Clearly we have then that lim n → + ∞ x n = + ∞ and lim n → + ∞ f ( x n ) 6 = y . Remark: This proof is analogous to the proof of Theorem 4.1.1. Just think in this case of + ∞ as a limit point of the domain of f . 4.1.5.5 Show that the function f ( x ) = x β on [0 , 1] for 0 < β ≤ 1 satisfies a H¨ older condition of order α for 0 < α ≤ β but not for α > β ....
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 Fall '07
 HUBBARD
 Math, Topology, Metric space, 4.1.5.4, 4.1.5.5, 4.1.5.7, 4.1.5.8

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