Solutions for Homework #6

1 Solutions for Homework #6 Math 451, Fall 2014, Section 3 1) Claim: If g(x) = x3 , then limx→1 Proof: Let h(x) = g(x)−1 x−1 . g(x)−1 x−1 = 3. Notice that R−{1} ⊂ dom(h) and R is an open interval containing 3 xn −1 n )−1 2 1. If (xn ) is a sequence in R − {1} converging to 1, then g(x xn −1 = xn −1 = xn + xn + 1 for all n ∈ N (since xn 6= 1). The multiplication limit law implies that lim x2n = 12 = 1, so then the addition limit law implies that lim x2n + xn = 1 + 1 = 2. Since lim x2n + xn = 2 and lim 1 = 1, the addition limit law implies that lim g(xn ) − 1 = lim(x2n + xn ) + 1 = 2 + 1 = 3. xn − 1 n )−1 So, if (xn ) is a sequence in R − {1}, then lim g(x xn −1 = 3, which implies that limx→1 g(x)−1 x−1 = 3. limx→1 x2n Remark: One may alternatively prove that + xn + 1 = 3 by using the fact that the 2 polynomial h(x) = x + x + 1 is continuous. We discussed this approach in class. 2) Claim: If [a, b] is a closed interval and f is a real-valued function which is continuous on [a, b], then f ([a, b]) = {f (x) | x ∈ [a, b]} is a closed set. Proof: Let (yn ) be a convergent sequence in f ([a, b]) and let y = lim yn . For all n ∈ N, there exists xn ∈ [a, b] such that f (xn ) = yn . Since (xn ) ⊂ [a, b], it is bounded, so the BolzanoWeierstrass Theorem implies that (xn ) has a convergent subsequence (xnk ). Let x = lim xnk . Since [a, b] is closed, x ∈ [a, b]. Since f is continuous on [a, b], lim f (xnk ) = f (x). But since (f (xnk )) = (ynk ) is a subsequence of the convergent sequence (yn ), lim f (xnk ) = lim yn = y, so f (x) = y, which implies that y ∈ f ([a, b]). Therefore, if (yn ) is a convergent sequence in f ([a, b]), then lim yn ∈ f ([a, b]), so f ([a, b]) is closed. Alternate Proof: Theorem 18.1 implies that there exists x0 , y0 ∈ [a, b] so that f (x0 ) ≤ f (x) ≤ f (y0 ) for all x ∈ [a, b], which implies that f ([a, b]) is contained in [f (x0 ), f (y0 )]. But Corollary 18.3 implies that f ([a, b]) is either an interval or a point, but since it is contained in the closed interval (or point) [f (x0 ), f (y0 )] and contains f (x0 ) and f (y0 ), we see that f ([a, b]) = [f (x0 ), f (y0 )]. Since we have previously established that [f (x0 ), f (y0 )] is closed, this implies that f ([a, b]) is closed. 19.4) a) Claim: If f is uniformly continuous on a bounded set S, then f is bounded on S. Proof: Suppose that f is not bounded on S. Then, for all n ∈ N, there exists xn ∈ S such that |f (xn )| > n. Since S is bounded, (xn ) is a bounded sequence. The Bolzano-Weierstrass Theorem implies that (xn ) has a convergent subsequence (xnk ). Since (xnk ) converges, Lemma 10.9 implies that (xnk ) is Cauchy. Theorem 19.4 implies that (f (xnk )) is Cauchy, and hence, by Lemma 10.10, bounded. So, there exists M > 0 such that |f (xnk )| ≤ M for all k. However, if k > M , then |f (xnk )| > nk ≥ k > M . Since we have found a contradiction, it must be the case that f is bounded on S. 1 19.5d) g(x) = x−3 is not uniformly continuous on (0, 3). Proof: Suppose that g is uniformly continuous on (0, 3). For all n ∈ N, let xn = 3− n1 ∈ (0, 3). Since lim n1 = 0 and lim 3 = 3, the subtraction limit law implies that lim xn = lim 3 − n1 = 3 + 0 = 3. Since (xn ) is convergent, it is Cauchy (by Lemma 10.9). Theorem 19.4 then implies that (g(xn )) is Cauchy, hence bounded (by Lemma 10.10). So, there exists M > 0 such that 2 |g(xn )| ≤ M for all n ∈ N. However, if n > M , then |g(xn )| = n > M , so we have obtained a contradiction. Therefore, g is not uniformly continuous on (0, 3). Alternate proof: Suppose that g is uniformly continuous on (0, 3). Then, since (0, 3) is bounded, exercise 19.4a implies that g is bounded on (0, 3). Therefore, there exists M > 0 such that |g(x)| ≤ M for all x ∈ (0, 3). Choose n ∈ N such that n > M . Then x = 3 − n1 ∈ (0, 3), but |g(x)| = n > M which gives a contradiction. Therefore, g is not uniformly continuous on (0, 3). √ 20.11b) Claim: If b > 0, then limx→b √ Proof: Let g(x) = √ x− b x−b . √ x− b x−b = 1 √ . 2 b Notice that (0, ∞) is an open interval containing b and (0, ∞) − √ √ x − b {b} ⊂ dom(g). If (xn ) is a sequence in (0, ∞) − {b} and lim xn = b, then g(xn ) = xnn −b = 1 √ for all n ∈ N (since xn 6= b for any n). Since lim xn = b, Example 8.5 implies that √ xn + b √ √ √ √ √ √ lim xn =√ b. Since lim b = b and lim x = b, the addition limit law implies that n √ √ 1 . lim xn + b = 2 b. The reciprocal limit law then implies that lim g(xn ) = lim √x 1+√b = 2√ b n We have √shown that whenever (x√n ) is a sequence in (0, ∞) − {b} such that lim xn = b, then √ lim xn − b xn −b = 1 √ , 2 b √ so limx→b x− b x−b = 1 √ . 2 b 20.16a) Suppose that L1 = limx→a+ f1 (x) and that L2 = limx→a+ f2 (x). Claim: If f1 (x) ≤ f2 (x) for all x ∈ (a, b), then L1 ≤ L2 Proof: Let (xn ) be a sequence in (a, b) which converges to a. Since limx→a+ f1 (x) = L1 , lim f1 (xn ) = L1 , and, since limx→a+ f2 (x) = L2 , lim f2 (xn ) = L2 . Our assumptions imply that f1 (xn ) ≤ f2 (xn ) for all n. Then, by the comparison lemma given on the problem sheet (and proven in class), L1 = lim f1 (xn ) ≤ lim f2 (xn ) = L2 as desired. (Notice that the comparison law stated on the problem sheet holds for L ∈ R, L = +∞ or L = −∞, by the proof given in class, if L ∈ R, and by Exercise 9.9, if not.) b) If f1 (x) < f2 (x) for all x ∈ (a, b), then it need not be the case that L1 < L2 Example: Let f1 (x) = x and f2 (x) = 2x and (a, b) = (0, 1). Then, f1 (x) < f2 (x) for all x ∈ (0, 1). Since f1 and f2 are continuous on R, lim f1 (x) = lim f1 (x) = f1 (0) = 0 = f2 (0) = lim f2 (x) = lim f2 (x) x→0+ x→0 x→0 x→0+ 20.17) Claim: If limx→a+ f1 (x) = limx→a+ f3 (x) = L and f1 (x) ≤ f2 (x) ≤ f3 (x) for all x ∈ (a, b), then limx→a+ f2 (x) = L. Proof: Let (xn ) be a sequence in (a, b) such that lim xn = a. Since limx→a+ f1 (x) = limx→a+ f3 (x) = L, lim f1 (xn ) = lim f3 (xn ) = L. Since f1 (xn ) ≤ f2 (x) ≤ f3 (xn ) for all n ∈ N and lim f1 (xn ) = lim f3 (xn ) = L, if L ∈ R, then the squeeze principle for sequences implies that lim f2 (xn ) = L. If L = +∞ or L = −∞ then Exercise 9.9 implies that lim f2 (xn ) = L. So, if (xn ) is a sequence in (a, b) such that lim xn = a, then lim f2 (xn ) = L, which implies that limx→a+ f2 (x) = L as claimed. 20.20b) Suppose that limx→aS f1 (x) = +∞ and limx→aS f2 (x) = L2 > 0. Claim: limx→aS (f1 f2 )(x) = +∞. Proof: Let (xn ) be a sequence in S which converges to a. Since limx→aS f1 (x) = +∞, lim f1 (xn ) = +∞. Since limx→aS f2 (x) = L2 , lim f2 (xn ) = L2 > 0. Theorem 9.9 implies that lim f1 (xn )f2 (xn ) = +∞, so lim(f1 f2 )(xn ) = +∞. We have shown that if (xn ) is a sequence in S which converges to a, then lim(f1 f2 )(xn ) = +∞, so limx→aS (f1 f2 )(x) = +∞.