Math 312 Intro to Real Analysis Homework 6 Solutions Stephen G Simpson Friday April 10 2009 The assignment consists of Exercises 17 3 a b c f 17 4 17 9 c d 17 10 a b 17 14 18 5 18 7 19 1 19 2 b c 19 5 in the Ross textbook Each exercise counts 10 points 17 3 a By Theorem 17 4 ii applied three times cos4 x is continuous Then by Theorem 17 4 i 1 cos4 x is continuous Then by Theorem 17 5 loge 1 cos4 x is continuous Since 1 cos4 x 1 for all x the domain of this function is the entire real line b Since x e log x it is clear by Theorem 17 4 that x is continuous It then follows as usual that sin2 x cos6 x is continuous Again the domain of this function is the entire real line 2 2 c We have 2x ex log 2 so as before this is continuous Again the domain is the entire real line f By Theorem 17 4 iii 1 x is continuous for all x 6 0 It then follows as usual that x sin 1 x is continuous for all x 6 0 17 4 Example 5 in 8 says that if lim xn x and xn 0 for all n then lim xn x According to Definition 17 1 this says precisely that x is continuous for all x 0 17 9 c Let f x x sin 1 x for x 6 0 and let f 0 0 We wish to show that f is continuous at 0 using the definition of continuity Given 0 we must find 0 such that f x whenever x We have f x x sin 1 x x since sin y 1 for all y So let Then clearly x implies x which implies f x Q E D d Let g x x3 We wish to show that g is continuous at x0 for an arbitrary x0 Given 0 and x0 we must find 0 depending on and x0 such that g x g x0 whenever x x0 For this particular function g x we have g x g x0 x3 x30 x2 xx0 x20 x x0 Even though we have not yet chosen our we may safely assume that 1 Then x x0 implies x x0 1 which implies 1 x2 xx0 x20 x0 1 2 x0 1 x0 x0 2 3 x0 2 3 x0 1 so let M 3 x0 2 3 x0 1 Then g x g x0 M x x0 so it will work to choose min 1 M Note that M depends on x0 so depends on both and x0 17 10 a We have f x 1 for all x 0 and f 0 0 Thus lim 1 n 0 but lim f 1 n 1 6 0 f 0 so f is disconinuous at 0 in view of Definition 17 1 b We have g x sin 1 x for all x 6 0 and g 0 0 Letting an n 21 we see that lim 1 an 0 but g 1 an sin an 1 n so lim g 1 an does not exist Thus g is discontinuous at 0 in view of Definition 17 1 17 14 Define f x 0 if x is irrational and f r 1 n for all rational numbers r m n where n 0 and GCD m n 1 We want to show that f is continuous at x if x is irrational and discontinuous at r if r is irrational We opt to use the definition of continuity Consider what happens near x when x is irrational Given 0 consider the set G consisting of all numbers r such that f r For all r G we have r m n for some n 1 Thus G is a discrete set of rational numbers Here discrete means that G contains only a finite number of points in each interval Consequently since x G we can find an open interval a b which contains x and is disjoint from G On this interval a b we have f z for all z in the interval Thus letting min x a x b we have 0 and f z whenever z x Since f x 0 we see that f is continuous at x On the other hand consider what happens near r when r is rational We have f r 1 n 0 Let f r and consider any 0 Clearly there are irrational numbers x in the open interval r x r and for these x s we have f x 0 Thus f x f r even though x r This shows that f is discontinous at r 18 5 a Let f and g be continuous on a b and assume f a g a and f b g b Consider the function h x f x g x Clearly h is continuous on a b and h a 0 and h b 0 By the Intermediate Value Theorem there is at least one x a b such that h x 0 Then clearly f x g x b Example 1 on page 128 may be viewed as the special case of part a where a b 0 1 and f x 0 1 and g x x for all x 0 1 18 7 Let f x x2x Clearly f x is continuous for all x and f 0 0 and f 1 2 Since 0 1 2 it follows by the Intermediate Value Theorem that f x 1 for some x 0 1 Clearly x 6 0 1 so x 0 1 19 1 a Uniformly continuous by Theorem 19 2 b Uniformly continuous by Theorem 19 2 2 c Uniformly continuous by part b d x3 is not uniformly continous because x3 x 3 3x2 3 whenever x 1 no matter how small is e 1 x3 is not uniformly continouous on 0 1 by Theorem 19 5 f sin 1 x2 is not uniformly continous on 0 1 by Theorem 19 5 g f x x2 sin 1 x is uniformly continouus on 0 1 by Theorem 19 5 because it can be extended to a continous function on 0 1 by defining f 0 0 19 2 b For x y 0 3 we have x2 y 2 x y x y 6 x y Consequently given 0 we have x2 y 2 whenever x y 6 and x y 0 3 Letting 6 we see directly that x2 is uniformly continuous on 0 3 in terms of the definition of uniform continuity c For x y 1 2 we have 1 1 x y x y 4 x y xy Consequently given 0 we have 1 1 x y whenever x y 4 and x y 1 2 Letting 4 we see directly that 1 x is uniformly continuous on 1 2 in terms of the definition of uniform continuity 19 5 a Uniformly continuous by Theorem 19 5 …