Midterm Exam Questions with Solution - Analysis 1 | MATH 3001, Exams of Advanced Calculus

Download Midterm Exam Questions with Solution - Analysis 1 | MATH 3001 and more Exams Advanced Calculus in PDF only on Docsity! NAME Math 3001 Midterm Examination 2 Page 1 Math 3001 Midterm 2 Solutions 1. (25 points) Prove that lim x→2 x+ 3 x− 1 = 5 using only the definition. Solution: The first step is to simplify f(x)− 5. We have f(x)− 5 = x+ 3 x− 1 − 5 = x+ 3− 5x+ 5 x− 1 = −4x+ 8 x− 1 = −4 x− 1 (x− 2). Taking absolute values we get |f(x)− 5| = 4|x−1| |x− 2|. We need to get a lower bound for |x − 1| in order to get an upper bound for 4|x−1| . Suppose first that |x − 2| < 1 2 ; then −1 2 < x − 2 < 1 2 and 1 2 < x − 1 < 3 2 . Thus |x − 1| > 1 2 , and 4 |x−1| < 8, assuming |x− 2| < 1 2 . So if |x− 2| < 1 2 , we have |f(x)− 5| < 8|x− 2|.. Now let ε > 0 be arbitrary, and choose δ = min{1 2 , ε 8 }. Then 0 < |x− 2| < δ implies |f(x)− 5| < 8|x− 2| < ε, as desired. 2. (25 points) Define a sequence (sn) by s1 = 1, sn+1 = 3 − 1 sn for n ∈ N. Prove that (sn) is bounded below by 1 and above by 3, and that it is monotone. To what does it converge? Solution: We prove these things by induction. To get it bounded below by 1, we notice that s1 = 1, and if sn ≥ 1 for some n, then 1sn ≤ 1 and − 1 sn ≥ −1, so that sn+1 ≤ 3− 1 sn ≥ 3− 1 = 2 > 1. Thus by induction we have sn ≥ 1 for all n ∈ N. To prove sn ≤ 3 for all n, notice that s1 = 1 ≤ 3. Now suppose sn ≤ 3 for some n ≥ 0; then 1 sn ≥ 1 3 and − 1 sn ≤ −1 3 , so that sn+1 = 3− 1 sn ≤ 3− 1 3 = 8 3 < 3, as desired. So by induction we get sn ≤ 3 for all n ∈ N. Finally to prove sn is monotone, we compute sn+2 − sn+1 = ( 3− 1 sn+1 ) − ( 3− 1 sn ) = 1 sn − 1 sn+1 = sn+1 − sn snsn+1 . NAME Math 3001 Midterm Examination 2 Page 2 Since sn ≥ 1 and sn+1 ≥ 1, we know sn+2−sn+1 and sn+1−sn always have the same sign. Now s2 = 3 − 1 = 2 > s1, so that the sequence is initially increasing; thus it is always increasing, and we always have sn+1 > sn. Since the sequence is increasing and bounded above, it must converge to a limit satisfying L = 3− 1 L , or L2 − 3L+ 1 = 0. The solutions of this quadratic equation are L = 3± √ 9− 4 2 = 3± √ 5 2 . We must choose the plus sign since the sequence starts at 1 and increases, and 3− √ 5 2 < 1. 3. (25 points) Let us say, here on this test and never again, that a function f : R → R has kablimit L at c ∈ R and write kab x→c f(x) = L if ∀ε > 0, ∀δ > 0, |x− c| < δ ⇒ |f(x)− L| < ε. Prove that there is exactly one function f with kab x→3 f(x) = 9. (In other words, prove that there is one, and also prove there cannot be any other.) Solution: The obvious candidate is f(x) = 9 for all x. To prove this has kablimit 9, let ε > 0 and δ > 0 be any numbers. Then if |x− 3| < δ, we have |f(x)− 9| = 0 < ε, as desired. To prove that there cannot be any other function satisfying this definition, suppose f is a function with f(a) = b for some a ∈ R and some b 6= 9. We want to prove the negation of the kablimit definition. The negation is that there exists ε > 0 and there exists δ > 0 such that there is an x ∈ R with |x− 3| < δ and |f(x)− 9| ≥ ε. Choose ε = |b − 9| > 0 by assumption. Choose δ = |a − 3| + 1 > 0. Choose x = a. Then |x− a| < δ by construction but |f(x)− 9| = |b− 9| ≥ ε, which is what we wanted. 4. (25 points) For each of the following, determine whether the statement is TRUE or FALSE. For full credit, you must JUSTIFY your answers, either by giving a brief argument or citing a result from the book or class, or by giving a counterexample. One sentence is enough for all of these. (a) Every sequence which is increasing and bounded above is Cauchy. TRUE. An increasing sequence which is bounded above must converge, and any con- vergent sequence must be Cauchy, by theorems from class. (b) If f : R → R and lim x→c f(x) = L, then for every sequence (sn) with lim n→∞ sn = c, we have lim n→∞ f(sn) = L. FALSE. This is not exactly the same as the sequence criterion for limits, which says that if sn 6= c for all n and lim sn = c, then lim f(sn) = L. As a counterexample, let f(x) = 0 for all x 6= 0 and f(0) = 1, wiht c = 0 and sn = 0 for all n; then limx→0 f(x) = 0 but lim sn = 0 and lim f(sn) = 1.