Download Answers to Midterm 2 - Loyola University Chicago

Midterm 1
Loyola University Chicago
Math 351, Fall 2010
Name (print):
Signature:
Please do not start working until instructed to do so.
You have 50 minutes.
You must show your work to receive full credit.
Calculators are OK (but don’t expect them to be useful).
You may use one double-sided 8.5 by 11 sheet of handwritten (by you) notes.
Problem 1.
Problem 2.
Problem 3.
Problem 4.
Problem 5.
Total.
Problem 1.(20 points) Calculate the limits. (I don’t expect proofs here, but some calculation justifying the answer is needed.)
a.(4 points)
3en + 5e2n − 7
n→∞ 9e2n + 11 − en
lim
Solution:
3en + 5e2n − 7 e−2n
5
3e−n + 5 − 7e−2n
=
lim
=
n→∞ 9e2n + 11 − en e−2n
n→∞ 9 + 11e−2n − e−n
9
lim
b.(5 points)
n2
n2
−
n→∞ n − 1
n+1
lim
Solution:
n2
n2 (n + 1)
n2 (n − 1)
2n2
n2
−
= lim
−
= lim 2
=2
n→∞ n − 1
n + 1 n→∞ (n − 1)(n + 1) (n + 1)(n − 1) n→∞ n − 1
lim
c.(4 points)
lim
n→∞
1
1−
3n
4
Solution:
lim
n→∞
d.(4 points)
1
1−
3n
4
=
1
1 − lim
n→∞ 3n
4
=1
54 − (−1)n
n→∞
n − 2n
lim
Solution:
1
54 − (−1)n
= lim (54 − (−1)n )
=0
n→∞
n→∞
n − 2n
n − 2n
1
because 54 − (−1)n is bounded and lim n−2
n = 0.
lim
e.(4 points)
1
lim (3n + 4n ) n
n→∞
1
1
1
1
1
Solution: (4n ) n < (3n + 4n ) n < (4n + 4n ) n so 4 < (3n + 4n ) n < 2 n 4 so
1
lim (3n + 4n ) n = 4
n→∞
Problem 2.(10 points total) Use the definition of the limit of a sequence to show that
3
3n2 − 1
=
2
n→∞ 5n + 4
5
lim
Solution: given any ε > 0, we must find N such
2
3n − 1
5n2 + 4 −
that for all n > N ,
3 < ε.
5
Do some algebra:
2
3n − 1 3 5(3n2 − 1) − 3(5n2 + 4) −17 17
17
=
<
=
5n2 + 4 − 5 = 2
2
2
5(5n + 4)
5(5n + 4)
5(5n + 4)
25n2
It is enough then to have
17
<ε
25n2
So, pick any N such that
r
17
<N
25ε
Problem 3.(10 points total) If a set S ⊂ R contains one of its upper bounds, prove that this upper
bound is the supremum of S.
Solution: let u be an upper bound of S and suppose that u ∈ S. To show that u is a supremum of
S, it is only left to show that u is the least upper bound. Let v be any other upper bound of S. Then
v ≥ s for all s ∈ S, and because u ∈ S, v ≥ u. Hence u is the least upper bound.
2
Problem 4.(10 points) Let x1 = 3 and xn+1 = xn + 4 for n ∈ N. Show that the sequence (xn ) is
3
monotone, bounded, and find its limit.
Solution: Try: x1 = 3, x2 = 6, x3 = 8. The sequence appears increasing. Let’s try proving it.
x1 < x2 by inspection. Now, given that xn−1 < xn , we have 23 xn−1 < 23 xn , 23 xn−1 + 4 < 23 xn + 4, and
thus xn < xn+1 . Mathematical induction implies that xn < xn+1 for all n ∈ N, and the sequence is
increasing.
It may not seem clear what upper bound to pick. However, if the sequence is convergent, then,
because it is increasing, the limit is an upper bound. Suppose for a moment that the limit exists.
Then limn→∞ = L, limn→∞ xn+1 = L, and
2
L= L+4
3
and so L = 12. So, we have a candidate for an upper bound: 12. Lets check: x1 < 12 by inspection.
If xn < 12 then 32 xn < 8, 23 xn + 4 < 12, and thus xn+1 < 12. Mathematical induction implies that
xn < 12 for all n ∈ N.
Now, we have a sequence that is increasing and bounded. Consequently, it has a limit. By now,
we know the limit must be 12.
Problem 5.(10 points) Recall the following definition:
Definition. The sequence (xn ) is convergent to L ∈ IR, denoted limn→∞ xn = L, if for every
ε > 0 there exists N such that for all n > N we have |xn − L| < ε.
Use this definition (and do not quote any limit theorems from the book or the lecture) to prove the
following theorem:
Theorem. Let (xn ) and (yn ) be convergent sequences, with limn→∞ xn = X, limn→∞ yn = Y .
Suppose that X > 0 and Y < 0. Prove that there exists a real number K such that for all
n > K, xn yn < 0.
Solution 1: Idea: show that xn are eventually positive, yn are eventually negative, so xn yn are
eventually negative. Details: in the definition of limn→∞ xn = X, consider ε = X/2; get that there
exists Nx such that for all n > N we have |xn − X| < ε and thus xn > X/2 and thus xn > 0. In the
definition of limn→∞ yn = Y , consider ε = −Y /2; get that there exists Ny such that for all n > N we
have |yn − Y | < ε and thus yn < Y /2 and thus yn < 0. Let N = max{Nx , Ny }. Then for all n > N ,
xn > 0, yn < 0, so xn yn < 0.
Solution 2: Idea: first prove that lim xn yn = lim xn lim yn = XY < 0. Second, prove that because
lim xn yn < 0, xn Yn < 0 for big enough n.