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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→∞ For every  > 0 there exists some M > 0 such that whenever x > M we have |f(x) − L| < .

(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 ∈ I and |x − y| < δ we have |f(x) − f(y)| < . Note: f is continuous on I is defined as: for each y ∈ I and  > 0 there is some δ > 0 such that whenever x ∈ I and |x − y| < δ we have |f(x) − f(y)| < . 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. ∞ P (a) True / False ak converges if and only if lim ak = 0. k=0 k→∞ If the converges, the 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.

∞ k P (−1) (b) True / False 2 is absolutely convergent. k=1 k P |(−1)k/k2| = 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 . √ (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, ∞), 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 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 ∞ as x → 0+). Also, f has no minimum value on I since it approaches f(1) = 1 from above but 1 6∈ 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 . P A conditionally convergent series is a convergent series (e.g., ak converges) which does not converge absolutely (i.e., P |ak| diverges). Examples of such series abound. A simple such example is the alternating harmonic series: ∞ X (−1)k k k=1 We have that P |(−1)k/k| = P 1/k is the harmonic series which diverges (it’s a p-series with p = 1) but the series itself converges by the 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 ∈ Q f(x) = 0 x ∈ R − Q

Recall that we can approach any a ∈ R via a sequence of rational numbers (say rk → a, rk ∈ Q). Likewise, we can approach via irrational numbers (say pk → a, pk ∈ 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 (not the ) to prove your results. Carefully check hypotheses! ∞ 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 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 ∞ ∞ 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). ∞ 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 ∞ 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 ∈ R. Let  > 0. Define δ = /3 (this is positive). Suppose that |x − y| < δ = /3 so that 3|x − y| < . Then |f(x) − f(y)| = |(−3x + 2) − (−3y + 2)| = | − 3x − (−3y)| = |(−3)(x − y)| = 3|x − y| < . Therefore, f(x) = −3x + 2 is continuous at y (for all y ∈ 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 √ (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 so√ we only consider√ √x’s above√ 0 – this also√ takes care√ of our domain issue: x > 0 guarantees x is in the domain of f(x) = x). Thus 0 < x < 2 so that | x − 0| = 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) ∈ Q) for every x ∈ 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 ∈ 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. For example, f(x) = x is uniformly continuous on R. However, f(x)f(x) = x2 is not uniformly continuous on R. Our assumption that I is a closed bounded interval is vital!

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