8 - Math 310 - hw 8 solutions 18.4, 18.6, 18.10; Friday, 6....

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Unformatted text preview: Math 310 - hw 8 solutions 18.4, 18.6, 18.10; Friday, 6. Nov 2009 18.4 Let S ⊂ R and suppose there is a sequence ( x n ) n in S that converges to a number x / ∈ S . Show that there exists an unbounded continuous function on S . We define f : S → R by f ( x ) := 1 x- x for x ∈ S. Then f is continuous, because it is the composition of the two continuous functions: x 7→ x- x and y 7→ 1 /y . We now verify that f is unbounded. Let M > 0 be given. Then since x n → x there is N such that for all n ∈ N , n > N implies that < | x n- x | < 1 M . Let n ∈ N and suppose that n > N . Then since | f ( x n ) | = 1 | x n- x | > 1 1 /M = M, f is unbounded on S . 18.6 Prove that x = cos x for some x in (0 , π 2 ). We suppose that the function x 7→ cos x is continuous on R . Let f be the R-valued function defined by f ( x ) = x- cos x for x ∈ R . Then f is continuous since it is the difference of continuous functions. Observe that f (0) =- 1 < < π 2 = f π 2 ....
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This note was uploaded on 03/26/2010 for the course MATHEMARIC 131a taught by Professor Janeday during the Spring '10 term at San Jose State.

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