Download Limits of Sequences and Functions: Relating the Two with l'Hopital's Rule and more Quizzes Advanced Calculus in PDF only on Docsity! Quiz # 4 Solutions There are two types of limits at∞ (or, more correctly, “limits as the input increases without bound”) in this class. The first type of limit is the limit of a function which is defined at all x > b (and possibly other places). If f is such a function, it may be possible to calculate the limit lim x→∞ f(x). We also consider functions which are only defined on positive integers. To avoid confusion, we write an rather than a(x). The letter n, and the use of subscripts, is a clue that we are playing a different game now. Given such a function (which we call a sequence), it may be possible to calculate the limit lim n→∞an. Note that these two concepts of limit, although very similar, are still distinct. For example, there is a theorem called “l’Hôpital’s rule” which says things about the first kind of limit. Unfortunately, l’Hôpital’s rule doesn’t directly say anything about the second type of limit. We have a trick, however, where we can sometimes relate the first kind of limit to the second kind. Given a sequence {an}, we might be able to come up with a function f defined for all real numbers x ≥ 1 such that f(n) = an for positive integers. So, even though f is defined on a much larger set than a, it happens to coincide with a. If we can find such an f, and limx→∞ f(x) = L (this is a limit of a function defined for all real x ≥ 1), then limn→∞ an = L (this is a limit of a sequence). If you have defined such an f, it may be possible to use l’Hôpital’s rule to evaluate limx→∞ f(x). This gives you a way to attack problems involving limits of sequences with l’Hôpital’s rule. But it is not possible to apply l’Hôpital’s rule to sequential limits directly—besides, if {an} is a sequence, it doesn’t make sense to define a derivative for a, at least not in the normal way. 1. (7 points) Compute, if it exists, lim n→∞ lnn3n . (Hint: Relate this limit to the old type of limit.) Solution: Set f(x) = ln x 3x . This is a function on all real numbers x > 0. Note that for a positive integer n, f(n) = lnn 3n . Now, limx→∞ ln x =∞ and limx→∞ 3x =∞, so by l’Hôpital’s rule, lim x→∞ ln x3x = limx→∞ 1/x 3 = lim x→∞ 13x = 0. Hence, lim n→∞ lnn3n = 0. 2. (3 points) Find a positive integer N such that∣∣∣∣ 1n − 0 ∣∣∣∣ < 1100 whenever n > N. (This is the ε condition for the limn→∞ an = 0 when ε = 1/100.) Solution: It is sufficient to choose any N ≥ 100. If N ≥ 100 and n > N, then∣∣∣∣ 1n − 0 ∣∣∣∣ = 1n < 1N ≤ 1100 . So, possible correct answers include (but certainly aren’t limited to) “N = 100”, “N = 101”, and “N = 290345”. 1
