hw1-4 - HW SOLUTIONS WEEKS 1-4 1. Disclaimer Even though...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HW SOLUTIONS WEEKS 1-4 1. Disclaimer Even though Ive made every effort to make these solutions clear and readable, I am only human and can make mistakes. If you find a mistake, please let me know ASAP so I can correct it. Also, I have taken liberty of omitting some details that arent necessary for me to repeat. However, if you dont understand a step I have omitted, ask for clarification. 2. Week 1 2.1. Problem 1. f is continuous everywhere except at x = 0. It should be clear that f is continuous on (- , 0) (0 , ). To see it isnt continuous at 0, let x n be a sequence converging to 0 in (0 , ). Then we have lim n f ( x n ) = lim n 1 / ( x n ) = 6 = f (0) Therefore f is not continuous at 0. 2.2. Problem 2. . Part (i): Let f ( x ) = x 3 , and fix x R . Given , we need to find an appropriate . It will suffice to look only on the interval I = [ x- 1 , x +1]. For all x I , we have the following inequality: | x 3- x 3 | = | x- x | | x 2 + xx + x 2 | | x- x | ( | x | 2 + | x || x | + | x | 2 ) | x- x | ( | x + 1 | 2 + | x + 1 || x | + | x | 2 ) = | x- x | C x where C x = | x + 1 | 2 + | x + 1 || x | + | x | 2 . Now let = min { 1 , /C x } . Then | x- x | < = | x- x | < 1 = | x 2 + xx + x 2 | < C x Then it follows that | x- x | < = | x 3- x 3 | | x- x | C x < C x ( /C x ) = Part (ii): By part (i) above, f ( x ) = x 3 is continuous on R . Then the function g ( x ) = x 6 + 1 is continuous on R , and in fact x 6 + 1 1 > 0 in R as well. Then by theorem 17.4, the function f/g is continuous on R . 2.3. problem 3. . Part (i): We already know f ( r ) = 0 for any rational r ( a, b ). Let ( a, b ) be an irrational number. By density of rational numbers, there exists a sequence r n of rational numbers converging to . Then, by continuity of f , we have f ( ) = lim n f ( r n ) = lim n 0 = 0 Therefore f ( ) = 0. Since ( a, b ) was arbitrary, we have f ( x ) = 0 for any x ( a, b ). Part(ii) Since foth f and g are continuous on ( a, b ), by theorem 17.4 the function h = f- g is continuous on ( a, b ). Furthermore, for any rational number r ( a, b ), 1 2 HW SOLUTIONS WEEKS 1-4 we have h ( r ) = f ( r )- g ( r ) = 0. Then we can apply part (i) form above to find that h ( x ) = f ( x )- g ( x ) = 0 for any x ( a, b ). Therefore f ( x...
View Full Document

This homework help was uploaded on 04/08/2008 for the course MATH 125A taught by Professor Fukuda during the Fall '07 term at UC Davis.

Page1 / 5

hw1-4 - HW SOLUTIONS WEEKS 1-4 1. Disclaimer Even though...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online