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Test #2 Name: Answer Math 3220 Test #2 March 20th, 2019 Name: Answer Key Be sure to show your work! 1. (14 points) Definitions (a) Let f : ! . Give the definition for: lim f(x) = L. R R x!1 For every > 0 there exists some M > 0 such that whenever x > M we have jf(x) − Lj < . (b) Let f : I ! R. Give the definition for: f is uniformly continuous on I. For every > 0 there exists some δ > 0 such that whenever x; y 2 I and jx − yj < δ we have jf(x) − f(y)j < . Note: f is continuous on I is defined as: for each y 2 I and > 0 there is some δ > 0 such that whenever x 2 I and jx − yj < δ we have jf(x) − f(y)j < . The difference is that in the definition of continuity δ is chosen after both and y are selected. On the other hand, in the definition of uniform continuity, δ is chosen after is selected (so our δ must work for all y's). 2. (25 points) Determine which of the following statements are true or false. Circle your answer. 1 P (a) True / False ak converges if and only if lim ak = 0. k=0 k!1 If the series converges, the limit of its terms must be zero (this is the nth-term test). However, the converse is false. For example, the harmonic series P 1=k diverges but 1=k ! 0. 1 k P (−1) (b) True / False 2 is absolutely convergent. k=1 k P j(−1)k=k2j = P 1=k2 is a convergent p-series (where p = 2 > 1). (c) True / False If lim f(x) = L, then lim f(1=x) = L. x→−∞ x!0− This is analogous to Proposition 3.6 part (ii). Essentially, at x ! −∞, 1=x ! 0 from the negative side. Thus these limits match. (d) True / False If both lim f(x) and lim g(x) do not exist, then lim (f(x) + g(x)) does not exist. x!a x!a x!a While convergent plus divergent is divergent and convergent plus convergent is convergent, it is possible that divergent 2 2 plus divergent is convergent or divergent. For example: limx!0 1=x and limx!0(−1=x ) both do not exist (they go to 2 2 ±∞). But limx!0 1=x + (−1=x ) = limx!0 0 = 0 exists! (e) True / False Let g be continuous at x = a and let f be continuous at x = g(a). It is possible that f ◦ g is not continuous at x = a. The composition of continuous functions is a continuous function. p (f) True / False f(x) = x − 4 is continuous at x = 4. Elementary functions are continuous on their domains. The domain of f is [4; 1), so f is continuous at any x ≥ 4. (g) True / False Let f be continuous on I = (a; b). Then f has a maximum and minimum value on I. The Extreme Value Theorem says that a continuous function on a closed bounded interval must attain its max and min values. However, here we have an open interval, so the conclusion of the EVT might not hold. For example, f(x) = 1=x is continuous on I = (0; 1). However, it has no maximum value (f approaches 1 as x ! 0+). Also, f has no minimum value on I since it approaches f(1) = 1 from above but 1 62 I. (h) True / False Let f be continuous with f(2) = −3 and f(5) = 9. Then f must have a root in I = (2; 5). This follows from the Intermediate Value Theorem. Since f(2) = −3 < y = 0 < 9 = f(5) and f is continuous, there must be some 2 < x < 5 such that f(x) = y = 0. (i) True / False f(x) = x2 is uniformly continuous on R. f(x) = x2 is continuous on R, but we showed in class that it is not uniformly continuous on R. (j) True / False A continuous function is uniformly continuous on I = [a; b]. This is Theorem 3.22. 1 3. (14 points) Examples requested. (a) Give an example of a conditionally convergent series. P A conditionally convergent series is a convergent series (e.g., ak converges) which does not converge absolutely (i.e., P jakj diverges). Examples of such series abound. A simple such example is the alternating harmonic series: 1 X (−1)k k k=1 We have that P j(−1)k=kj = P 1=k is the harmonic series which diverges (it's a p-series with p = 1) but the series itself converges by the alternating series test (the terms alternate in sign and their magnitudes 1=k form a strict monotone sequence which converges to 0). (b) Give an example of a function f : R ! R such that f is not continuous anywhere. Creating a function that fails to be continuous anywhere is a little trick. We gave an example of such a function in class, in particular: Dirichlet's function. ( 1 x 2 Q f(x) = 0 x 2 R − Q Recall that we can approach any real number a 2 R via a sequence of rational numbers (say rk ! a, rk 2 Q). Likewise, we can approach via irrational numbers (say pk ! a, pk 2 R − Q). If f were continuous at a, we would have both f(rk) = 1 ! 1 = f(a) and f(pk) = 0 ! 1 = f(a) (contradiction). Thus f is not continuous anywhere. 4. (16 points) Use the direct comparison test (not the limit comparison test) to prove your results. Carefully check hypotheses! 1 X 2k (a) Prove that converges. k2 + 5k k=1 2k 2k For large k's, we have ≈ = (2=5)k. Thus this series is similar to a geometric series with r = 2=5 < 1, so it k2 + 5k 5k should converge. Let's compare. 2k 2k 2k 2k First, notice that 0 ≤ for all k ≥ 1. Next, ≤ = since shrinking a denominator grows a fraction. k2 + 5k k2 + 5k 5k 5 1 1 X 2k X Therefore, by the comparison test, converges because (2=5)k converges (this is a geometric series with k2 + 5k k=1 k=1 r = 2=5 < 1). 1 X k2 (b) Prove that diverges. 5k3 − k − 1 k=1 k2 k2 1 For large k's, we have ≈ = . Thus this series is similar to the harmonic series, so it should diverge. 5k3 − k − 1 5k3 5k Again, let's compare. k2 First, notice that ≥ 0 for k ≥ 1. The subtraction should worry us, but 5k3 − k − 1 ≥ 5k − k − k = 3k > 0 for 5k3 − k − 1 k2 k2 1 1 k ≥ 1. Next, ≥ = · since growing a denominator shrinks a fraction. Therefore, by the comparison 5k3 − k − 1 5k3 5 k 1 X k2 test diverges because 1=5 times the harmonic series diverges. 5k3 − k − 1 k=1 5. (16 points) Prove the following statements using -δ definitions. Do not call on limit/continuity theorems. (a) Prove that f(x) = −3x + 2 is continuous (everywhere). Pick some y 2 R. Let > 0. Define δ = /3 (this is positive). Suppose that jx − yj < δ = /3 so that 3jx − yj < . Then jf(x) − f(y)j = j(−3x + 2) − (−3y + 2)j = j − 3x − (−3y)j = j(−3)(x − y)j = 3jx − yj < . Therefore, f(x) = −3x + 2 is continuous at y (for all y 2 R). Note: Our choice of δ does not depend on y. Thus our proof shows that f is not only continuous but also uniformly continuous on R. 2 p (b) Prove that lim x = 0. x!0+ Let > 0. Define δ = 2 (this is positive). Suppose 0 < x < δ = 2 (notice that we are considering a right-handed limit sop we only considerp px's abovep 0 { this alsop takes carep of our domain issue: x > 0 guarantees x is in the domain of f(x) = x). Thus 0 < x < 2 so that j x − 0j = x < . This shows that the limit is 0. 6. (15 points) Choose one of the following statements to prove. I. Let f : R ! R be a continuous function such that f(x) is rational (i.e., f(x) 2 Q) for every x 2 R. Prove that f must be a constant function. Suppose that f is continuous and only takes on rational values. For sake of contradiction, suppose that f is not constant. Therefore, there exists some a; b 2 R with a < b and f(a) 6= f(b). We know that between any two real numbers there are (infinitely many) irrational (and rational) numbers. Let y be some irrational number between f(a) and f(b). Then by the Intermediate Value Theorem (this theorem applies since f is continuous), there is some a < x < b such that f(x) = y. But f(x) = y is irrational. This contradicts the fact that f only takes on rational values (contradiction). Therefore, f must be constant. II. Let f(x) and g(x) be uniformly continuous on I = [a; b]. Prove that f(x)g(x) is uniformly continuous on I. Here we just string a few theorems together. f and g are uniformly continuous on I so they are continuous on I (uniform continuity implies continuity). Therefore, fg is continuous on I (the product of continuous functions is a continuous function). Therefore, fg is uniformly continuous (if a function is continuous on a closed bounded interval: I = [a; b], then it is uniformly continuous). Note: In general, it is not true that the product of uniformly continuous functions is continuous.
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