hw3sol

# hw3sol - Math 125A Fall 2009 Solutions for Homework...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 125A, Fall 2009 Solutions for Homework 3 (Prepared by Matt Low) 1. Let ε > 0 be given. By the definition of uniform continuity, there exists a number δ > 0 with the property that if | a- b | < δ then | f ( a )- f ( b ) | < ε . Because | x n- y n | → 0, we may choose N so that | x n- y n | < δ for n ≥ N . Thus, if n ≥ N , then | x n- y n | < δ , and so | f ( x n )- f ( y n ) | < ε . 2. (a) Note that f consists of sums and compositions of continuous functions. Per Theorems 17.4 and 17.5, this means that f itself is continuous. It follows from Theorem 19.2 that f is uniformly continuous. (b) Define two sequences ( x n ) , ( y n ) by x n = 1 n and y n = 1 n 2 . It is clear that | x n- y n | → 0, but: lim n →∞ | f ( x n )- f ( y n ) | = lim n →∞ ln n = ∞ Because | f ( x n )- f ( y n ) | 6→ 0, this contradicts the property derived in Problem 1, and so we conclude that f is not uniformly continuous. Or: Since lim x → 0+ ln( x ) =-∞ , which is not a finite number, one cannot extend ln( x ) to be a continuous function on [0 , 1]. By the continuous extension theorem (theorem 19.5), ln1]....
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

hw3sol - Math 125A Fall 2009 Solutions for Homework...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online