Math 137 Online Week 2 Outline and Practice Problems
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Math 137 Online
Week 2 Outline and Practice Problems
Practice Problems
1. Use the formal definition of limits to prove each statement below:
2n
n→o n + 1
6n - 3n2 - 2
n→o (n - 1)2
(c) lim 1 - 2n = -< n→o
2. Determine if the following statements are true or false. If true, argue your case mathematically, if false, provide a counterexample.
(a) If lim an = lim bn = < then lim (an + bn ) = <
n→o n→o n→o
(c) If an ● bn ● cn for all n, lim an = L and lim cn = M, then lim bn = K with L ● K ● M .
an = ,
when n is a perfect square
otherwise
Use the definition of convergence to show this sequence does not have a limit of 0. Hint: consider building a contradiction.
4. Show that if a sequence /an { converges to L then there are infinitely many terms of the sequence that can be made arbitrarily close to one another. Specifically, show that eventually }an - am } can be made arbitrarily small (i.e. for a n, m passed a certain point). Note that m and n are not necessarily consecutive integers. Such sequences are called Cauchy Sequences.
5. In this question we will prove the following:
Assuming lim an = L, lim bn = M and an < bn for all n, then it must be the case that L < M
(a) Use the definition of limits to show that for all c, eventually
L - c < an < bn < M + c
(the word “eventually” is meant to take the place of the statement “for all n greater than some N”)
(b) Since c is arbitrary the inequality above might help you “feel” that L < M . That is, we can
make c so small that “basically L < M” . One way of showing this mathematically is to assume that L > M and come up with a contradiction.
That is, let L = M + d for some positive number d. Use this to arrive at a contradiction and thus deduce L < M .
6. Prove (using the definition) that if an > 0 for all n and lim an = L, then lim ^an = ^L. [Hint:
Consider the cases L = 0 and L 0 separately.]
Practice Quiz
1. Let /an { be a sequence such that lim an = <. Which of the following must be true?
(a) Every term of /an { is positive.
(b) No subsequence of /an { converges.
(c) lim 3n2 an = <
an
(e) lim -an = -<
(f) None of the above
2. Which cutoff values of N will guarantee that an = 2 + n2 is greater than M , assuming M > 5 (select all that apply):
(a) N = ^M
(b) N = ^M - 5
(c) N =
(d) N = M
(e) N = M2 + 2
(f) None of the above
2023-02-11