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

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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...
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## 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.

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

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