Advanced Calculus I Chapter 2 & 3 Homework Solutions October 30, 2009 2. Define f : ( 2, 0) R by f(x) = 2x2 + 3x 2. Prove that f has a limit at 2 and x + 2 find it. Note that when x 2 we have f(x) = 2x2 + 3x 2 x + 2 = 2x 1. Because the one value of x we never have to consider when talking about the limit at 2 is exactly 2 itself, it suffices to show that g(x) = 2x 1 has a limit at 2. We will guess that this limit is 5. To this end, let ǫ > 0 be given. We seek δ > 0 so that (2x 1) ( 5) < ǫ whenever 0 < x ( 2) < δ. Set δ = ǫ/2. Then we have as required. 0 < x + 2 < δ 2x + 4 < ǫ (2x 1) ( 5) < ǫ, 3. Give an example of a function f : (0, 1) R that has a limit at every point of (0, 1) except 1. Use the definition of limit of a function to justify the example. 2 Perhaps the simplest example is the following: { 0 if x (0, 1 2 f(x) = ) 1 if x [ 1, 1) 2 Suppose this function has a limit at 1/2 and call it L. Set ǫ = 1 3. Then there is some δ > 0 so that Choose x 1 and x 2 with Then 0 < x 1 2 < δ implies f(x) L < 1 3. 1 2 δ < x 1 < 1 2 < x 2 < 1 2 + δ. 1 = f(x 2 ) f(x 1 ) = f(x 2 ) L + L f(x 1 ) f(x 2 ) L + f(x 1 ) L < 2 3, a contradiction. Thus f has no limit at 1/2. I ll leave it as an exercise that f does have a limit at all other x (0, 1). (Hint: δ = ǫ.)
5. Suppose f : D R with x 0 an accumulation point of D. Assume L 1 and L 2 are limits of f at x 0. Prove L 1 = L 2. (Use only the definition; in later theorem, this uniqueness is assumed.) We show that L 1 = L 2 in essentially the same way we showed that the function in the problem above has no limit at 1/2: if L 1 L 2, then there s a gap between them, and we can choose ǫ > 0 small enough to prevent the function form being arbitrarily close to both of them. To this end, set ǫ = 1 3 L 2 L 1. Because L 1 is a limit at x 0, there is some δ 1 > 0 so that f(x) L 1 < ǫ whenever x D, 0 < x x 0 < δ 1. Similarly, there is some δ 2 > 0 so that f(x) L 2 < ǫ whenever x D, 0 < x x 0 < δ 2. Set δ = min{δ 1, δ 2 }. Then for any x D with 0 < x x 0 < δ, we have L 2 L 1 = L 2 f(x)+f(x) L 1 f(x) L 1 + f(x) L 2 < 2ǫ = 2 3 L 2 L 1. This is only possible if L 2 L 1 = 0, so that L 1 = L 2. 10. Consider f : (0, 2) R defined by f(x) = x x. Assume that f has a limit at 0 and find that limit. (Hint: Choose a sequence {x n } n=1 converging to 0 such that the limit of the sequence {f(x n )} n=1 is easy to determine.) Consider the sequence x n = 1 n. Then f(x n ) = ( ) 1/n 1 = 1 n n. n Assuming you know that lim n n = 1, it follows that lim f(x n ) = 1. If we further assume that f has a limit at zero, then this limit is 1. 11. Suppose f, g, and h : D R where x 0 is accumulation point of D, f(x) g(x) h(x) for all x D, and f and h have limits at x 0 with lim x x0 f(x) = lim x x0 h(x). Prove that g has a limit at x 0 and lim f(x) = lim g(x) = lim h(x). x x 0 x x0 x x0
Set L equal to the limit of f and h at x 0, and let ǫ > 0 be given. Then there is some δ 1 > 0 so that f(x) L < ǫ whenever x D, 0 < x x 0 < δ 1. Similarly there is δ 2 > 0 so that h(x) L < ǫ whenever x D, 0 < x x 0 < δ 2. Set δ = min{δ 1, δ 2 }. Then for x D with 0 < x x 0 < δ we have as required. L ǫ < f(x) g(x) h(x) < L + ǫ, Note that this result also follows from problem 20 below. 15. Let f : D R with x 0 as an accumulation point of D. Prove that f has a limit at x 0 if for each ǫ > 0, there is a neighborhood Q of x 0 such that, for any x, y Q D, x x 0, y x 0, we have f(x) f(y) < ǫ. Let ǫ > 0 be given. Then by assumption, there is some neighborhood Q of x 0 so that f(x) f(y) < ǫ whenever x, y Q D with x x 0 y. Because x 0 is an accumulation point of D, we may choose a sequence {x n } n=1 of points in D converging to x 0. Thus there is some N so that x n Q D whenever n N. Thus given ǫ > 0, we have found N so that n, m N x n, x m Q D f(x n ) f(x m ) < ǫ. In other words, the sequence {f(x n )} n=1 is Cauchy, and hence converges. The result now follows from Theorem BLAH. 18. Define g : (0, 1) R by g(x) = 1 + x 1. Prove that g has a limit at 0 and find it. x Using some calculus, we guess that lim x 0 g(x) = 1. To prove this, 2 let ǫ > 0 be given. We seek δ > 0 so that g(x) 1/2 < ǫ whenever 0 < x < δ. Choose δ = ǫ (in fact, we could get away with δ = 8ǫ). Suppose 0 < x < δ with 0 < x < 1. Then g(x) 1/2 = 1 + x 1 1 x 2 = 2 1 + x 2 x 2x
= 2 1 + x (2 + x) 2x 2 1 + x + (2 + x) 2 1 + x + (2 + x) = x 2 2x(2 1 + x + (2 + x)) < x 8 < x < δ = ǫ. (For the first inequality in the last displayed line, we use the fact that x > 0, so that 1 + x > 1 and 2 + x > 2 to get the 8 in the denominator.) 20. Prove Theorem 2.5: Suppose f : D R and g : D R, x 0 is an accumulation point of D, and f and g have limits at x 0. If f(x) g(x) for all x D, then lim f(x) lim g(x). x x 0 x x0 Let {x n } n=1 be a sequence of points in D coverging to x 0. Then {f(x n )} n=1 converges to lim x x0 f(x), and {g(x n )} n=1 converges to lim x x 0 g(x). Applying a previous result about sequences gives lim f(x) = lim {f(x n )} lim {g(x n )} = lim g(x). x x 0 x x0 22. Show by example that, even though f and g fail to have limits at x 0, it is possible for f + g to have a limit at x 0. Give similar examples for fg and f g. Suppose f(x) is any function that does not have a limit at x 0. Then g(x) = f(x) also fails to have a limit at x 0. On the other hand, (f + g)(x) = 0 for all x, and so has a limit at x 0. Similarly, if one assumes that f(x) 0 for all x, then f not having a limit at x 0 implies neither does g(x) = 1. On the other hand, f(x) (fg)(x) = 1 for all x, and so has a limit at x 0. Finally, if f(x) has no limit at x 0 and is never zero, then f f (x) = 1 for all x, and so has a limit at x 0. For all these to work, we need only know that there are functions that are never zero that fail to have limits at prescribed points. These are actually rather easy to come by. Here s an example that s far worse than necessary. Define { 1 if x Q f(x) = 1 if x / Q Then f(x) is never zero, and f fails to have a limit at any point. 24. Let f : [a, b] R be monotone. Prove that f has a limit both at a and at b.
Assume without loss that f is increasing. We first show that f has a limit at b. Because f is increasing, f(b) is an upper bound for B = {f(x) x < b}. It follows that this set has a least upper bound (a sup). Denote this by L = sup B. (This is what is called L(b) in the proof from class about monotone functions.) We will show that lim x b f(x) = L. Let ǫ > 0 be given. Then because U is the least upper bound of B, it follows that L ǫ is not an upper bound for B at all. Thus there is some x 0 < b so that L ǫ < f(x 0 ) L. Moreover, because f is increasing, we know that L ǫ < f(x 0 ) f(x) L whenever x 0 < x < b. Thus if we take δ = b x 0, then f(x) L < ǫ whenever 0 < x b < δ, as required. The argument for a, as well as for decreasing f, is similar. 25. Suppose f : [a, b] R and define g : [a, b] R as follows: g(x) = sup{f(t) a t x}. Prove that g has a limit at x 0 if f has a limit at x 0 and lim t x0 f(t) = f(x 0 ). 1. Define f : R R by f(x) = 3x 2 2x + 1. Show that f is continuous at 2. We will do this two ways. First we use the theorem that f is continuous at 2 if and only if lim x 2 f(x) = f(2). Let ǫ > 0 be given. We need to show that lim x 2 f(x) = 9. For this we need δ > 0 so that f(x) 9 = 3x 2 2x + 1 9 = 3x + 4 x 2 < ǫ when x 2 < δ. We can make x 2 as small as we like by choosing δ small. To bound 3x + 4, note that x 2 < 1 1 < x < 3 7 < 3x + 4 < 13 3x + 4 < 13. Thus assuming x 2 < 1 ensures 3x + 4 < 13. So if we also assume x 2 < ǫ/13, the product of the two will be less than ǫ. Here s the formal proof: Let ǫ > 0 be given, and set δ = min{1, ǫ/13}. Then x 2 < δ implies 3x + 4 < 13, so (3x 2 2x + 1) 9 = 3x + 4 x 2 < 13(ǫ/13) = ǫ when x 2 < δ.
6. Prove that f(x) = x is continuous for all x 0. 7. Suppose f : R R is continuous and f(r) = r 2 for each rational number r. Determine f( 2) and justify your conclusion. Let {x n } n=0 be a sequence of rational numbers converging to 2 (one may take x n to be the decimal expansion of 2 truncated at n decimal places, for instance). The because f is continuous, we know that f( 2) = lim {f(x n )} = lim {x 2 n} = ( lim {x n} )( ) lim {x n} = ( 2) 2 = 2. Note that the third equality is a result of the fact that, for sequences, the limit of the products is the product of the limits. 9. Define f : (0, 1) R by f(x) = sin 1. Can one define f(0) to make f continuous at 0? x Explain. No. Take ǫ = 1. Then for any δ > 0, there are points x 1 and x 2, with 0 < x 1, x 2 < δ so that f(x 1 ) = 1 and f(x 2 ) = 1. In particular, for any potential value f(0) we have 2 = f(x 1 ) f(x 2 ) = f(x 1 ) f(0)+f(0) f(x 2 ) f(x 1 ) f(0) + f(x 2 ) f(0). Thus it is impossible that both f(x 2 ) f(0) and f(x 1 ) f(0) both be less than ǫ = 1. 11. Define f : R R by f(x) = 8x if x is rational and f(x) = 2x 2 + 8 is x is irrational. Prove from the definition of continuity that f is continuous at 2 and discontinuous at 1.