7 - Math 310 - hw 7 solutions Wednesday, 28 Oct 2009 17.4,...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 310 - hw 7 solutions 17.4, 17.9d, 17.10c; Wednesday, 28 Oct 2009 17.4 Prove that the function x 7→ x is continuous on its domain [0 , ). Hint: Apply Example 5 in § 8. Let x 0 [0 , ) and let ( x n ) n be a sequence in [0 , ) such that lim n →∞ x n = x 0 . Then by Example 5 in § 8, lim n →∞ x n = x 0 . Hence, the function is continuous at x 0 by Definition 17.1. Since x 0 [0 , ) was chosen arbitrarily, the function x 7→ x is continuous on its domain [0 , ). ± 17.9d Prove that the following function is continuous at x 0 by verifying the ε - δ property of Theorem 17.2. g ( x ) = x 3 , x 0 arbitrary. Hint: x 3 - x 3 0 = ( x - x 0 )( x 2 + xx 0 + x 2 0 ). Let ε > 0 and let x 0 be given. Set M = 1 + | x 0 | . If | x - x 0 | < 1 we have: | x 0 | , | x | < M and x 2 0 , x 2 , | xx 0 | < M 2 . So we set δ := min ( 1 , ε 3 M 2 ) . Let x R and suppose that | x - x 0 | < δ ; then | x - x 0 | < 1 and so the above inequalities hold and therefore: | x 3 - x 3 0 | = | x 2 + xx 0 + x 2 0 || x - x 0 | ≤ ( x 2 + | xx 0 | + x 2 0 ) | x - x 0 | < 3 M 2 δ ε. Hence, | x 3 - x 3 0 | < ε and so g is continuous at x 0 by Theorem 17.2. ± 17.10c Prove that the function sgn (defined below) is discontinuous at the point x 0 = 0. You may use either Definition
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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.

Ask a homework question - tutors are online