Math 115 Spring 2014
Written Homework 2
Due Monday, February 10
Instructions: Write complete solutions on separate paper (not spiral bound). If multiple
pieces of paper are used, they must be stapled with your name and lecture written on each
page. In general, answers with no supporting work will not receive full credit. Please review
the Course Information document for more complete instructions.
1. The graph of f is given. Draw the graphs of the following functions.
(a) y = f (−x)
Solution: This is a reflection of the graph of f about the y-axis.
(b) y = f (x + 4) − 2
Solution: The term f (x + 4) causes a horizontal shift of the original graph to the left 4.
Then, the term −2 causes a vertical shift down of two units to the graph of y = f (x + 4).
(c) y = − 12 f (x) + 3
Solution: The term 12 f (x) causes a compression of the original graph.
Then, the scalar −1 causes a reflection about the x-axis to the graph of y = 21 f (x).
Then, the +3 causes a vertical shift up three units to the graph of y = − 12 f (x)
(d) y = 2f (x − 3)
Solution: The coefficient 2 causes a rescaling/stretch by a factor of 2.
Then, the f (x − 3) causes a horizontal shift of the graph to the right 3.
2. Consider the sequences defined by the following functions. For each one (i) find the first
5 terms, the 500th term, and the 5000th term of the sequence, (ii) determine the limit of
each sequence and (iii) list all of the following terms that correctly describe the sequence:
alternating, bounded from above, bounded from below, strictly increasing, strictly decreasing, convergent, divergent.
(a) f (n) :=
n−1
4n + 1
Solution:
(i) f (1) =
f (2) =
1−1
0
= =0
4(1) + 1
5
2−1
1
=
4(2) + 1
9
2
3
4
, f (4) = , and f (5) =
13
17
21
499
5000 − 1
4999
500 − 1
=
and f (5000) =
=
.
f (500) =
4(500) + 1
2001
4(5000) + 1
20001
f (3) =
(ii)By inspection, you may be able to see that the terms of this sequence are getting closer
499
500
4999
5000
and closer to 14 ( 2001
is nearly 2000
= 41 and 20001
is nearly 20000
= 14 ). But we also learned a
more formal way to compute this limit in lecture using the limit laws:
lim f (n) =
n→∞
=
=
=
=
n−1
n→∞ 4n + 1
n − 1 1/n
lim
·
n→∞ 4n + 1 1/n
1 − 1/n
lim
n→∞ 4 + 1/n
1−0
since we know that 1/n → 0 as n → ∞
4+0
1
4
lim
(iii) This sequence is bounded from above and bounded from below because all of the values
are clearly between 0 and 1; the sequence is strictly decreasing because each term is less than
1
the previous one; the sequence is convergent since it is approaching a single, finite value ( ).
4
(−1)n
(b) f (n) :=
3n
Solution:
(i)f (1) =
(−1)1
1
=−
3(1)
3
(−1)2
1
f (2) =
=
3(2)
6
1
1
1
f (3) = − , f (4) = , and f (5) = −
9
12
15
1
1
1
1
f (500) = (−1)500
=
and f (5000) = (−1)5000
=
.
3(500)
1500
3(5000)
15000
(ii) By inspection, you should notice that the absolute value of the terms is approaching
0. Also, we know from lecture that an = 1/n → 0 as n → ∞. Since 1/3 and −1/3 are
1
1 1
1
constants, we also know from the limit laws that ± an = · → ± · 0 = 0.
3
3 n
3
n
(−1)n
1
=
·
,
then
f
(n)
→
0
as
n
→
∞
or
lim
f
(n)
= 0.
Since f (n) := (−1)
2n
2
n
n→∞
(iii) This sequence is bounded from above and bounded from below since all the terms
are clearly between -1 and 1; the sequence is alternating since the signs alternate between
negative and positive; the sequence is convergent since the terms are approaching 0.
(c) an :=
n
−6
2
Solution:
(i)f (1) =
1
11
−6=−
2
2
2
− 6 = −5
2
7
9
f (3) = − , f (4) = −4, f (5) = −
2
2
500
5000
f (500) =
− 6 = 244, f (5000) = −
− 6 = 2494
2
2
f (2) =
(ii) By inspection, we see that the terms in this sequence will continue to increase without
bound as n goes to infinity. Hence the limit does not exist. That said, a more accurate
answer is
lim an = ∞,
n→∞
as this indicates that the terms in the sequence increase without bound forever.
(iii) This sequence is bounded from below since all the terms are greater than or equal to
11
− , strictly increase since each term is greater than the previous one, and divergent since
2
the terms do not approach a single, finite value.
(d) an := 6
Solution:
(i) This sequence is a constant sequence. Every term in the sequence is 6.
(ii) Since the terms in this sequence never vary from 6, in the long run the terms will always
be 6. Hence, 6 is the limit.
lim an = 6
n→∞
(iii) This sequence is bounded from above and bounded from below and convergent.
(e) f (n) := (−1)n (2n)
Solution:
(i) f (1) = (−1)1 (2(1)) = −2
f (2) = (−1)2 (2(2)) = 4
f (3) = −8, f (4) = 16, and f (5) = −32
f (500) = (−1)500+1 (2(500)) = (−1)501 (1000) = −1000
f (5000) = (−1)5000+1 (2(5000)) = (−1)5001 (10000) = −10000
(ii) Here, we need to be careful in our answer. The problem with just picking a number
of “large” numbers (500 and 5000 respectively) to evaluate the long run behavior of this
sequence is that we miss the alternating nature of this sequence. Note that for every even
number (−1)n = 1, but for ever odd number (−1)n = −1. For this sequence, the “size” or
“magnitude” (specifically, |f (n)| = |(−1)n (2n)| = 2n) increases without bound. But each
term in this sequence alternates. Thus, the limit of this sequence does not exist.
(iii) This sequence is alternating since the signs alternate between negative and positive and
it is divergent since the terms do not approach a single, finite value.
3. Find a formula for the general term an of each sequence, assuming that the pattern of
the first few terms continues.
1 2 3 4
(a)
, , , , ...
4 9 16 25
Solution: Here a1 = 14 , a2 = 29 , a3 =
3
,
16
and a4 =
4
.
25
We see that the numerators of each
n
successive terms increases by one. In fact, a generating formula must take the form .
?
Examining the denominator, we see that these are the squares of the integers: 22 , 32 , 42 ,
52 . In relation to n, the denominator of an is (n + 1)2 . Thus, a generating formula must take
n
the form (n+1)
2 . Hence,
n
an := (−1)n
.
(n + 1)2
2 3 4 5
(b) 1, − , , − , , ...
3 5 7 9
Solution: Here, we again see that the numerator is just n. The denominator is the set of
odd numbers. We know from lecture that the odd numbers are generated by the sequence
on := 2n − 1.
Lastly, we see that our sequence is an alternating sequence starting with the value 1. We
know that either (−1)n or (−1)n+1 must be used to generate the alternating pattern. Starting
from n = 1, (−1)n creates a −, +, −, +, . . . sequence and (−1)n+1 creates a +, −, +, −, . . .
sequence. Here we need to start with a positive term. Thus, a generating formula for this
sequence is
n
.
an := (−1)n+1
2n − 1
1
4. Let an :=
2n
(a) Find the first 10 terms of the sequence and graph the points (n, an ) on the Cartesian
plane.
Solution:
(b) Find the limit of the sequence, if it exists. Justify your answer, referencing the graph
you made in part (a).
Solution: Graphically, we see that the points on the graph will continue to get closer and
closer to the horizontal axis (y = 0). Hence, the limit of this sequence is 0.
1
= 0.
n→∞ 2n
lim
5. Suppose {an } is a strictly decreasing sequence.
(a) What is true about 2 consecutive terms, an and an+1 ?
Solution: Strictly Decreasing means that the next term must always be smaller than the
proceeding term. In particular,
an+1 < an .
(b) If an := 10 − 5n, prove that the sequence is strictly decreasing.
Solution: We need to show that an+1 < an . Note that for any n, an = 10 − 5n.
an+1 =
=
=
=
10 − 5(n + 1)
10 − 5n − 5
(10 − 5n) − 5
an − 5
Hence, an > an+1 always.
Note, we realize many of you approach this problem differently. You ”start” with an+1 <
an and hope to prove something true. While this is not an incorrect approach, we see many
make algebraic mistakes and ultimately prove something false. (Especially if your original
supposition was false.) We suggest the following approach.
an+1
10 − 5(n + 1)
10 − 5n − 5
5 − 5n
5
Since 5 < 10, we have shown an+1 < an .
?
?
?
?
?
an
10 − 5n
10 − 5n
10 − 5n
10
6.
(a) Give an example of an infinite sequence {an } which is bounded from below and bounded
from above, but lim an does not exist. You may give the sequence by listing the first several
n→∞
terms or by stating its generating function.
Solution: Using an example from the course notes, the function a(n) := (−1)n , which generates the sequence −1, 1, −1, 1, −1, ..., answers the question. The terms are bounded in the
interval [−1, 1], but the limit does not exist since the limit never tends towards a unique
limiting value.
(b) Give an example of an infinite sequence {an } which is strictly increasing, but
lim an 6= ∞. You may give the sequence by listing the first several terms or by stating its
n→∞
generating function.
1
Solution: Using an example from course notes, the function a(n) := , which generates
n
the sequence {1, 12 , 31 , 14 , 15 , ...}, is a strictly decreasing function. We can make the sequence
1
increasing by multiplying each term by −1. the function a(n) := − , which generates the
n
sequence {−1, − 21 , − 13 , − 41 , − 51 , ...}, is an example. The terms are strictly increasing, yet
bounded from below and bounded from above; the sequence has a limit of 0.
7. Consider the sequence generated by an :=
2n
.
n+1
(a) As we did in lecture, use the Limit Laws to determine the limit of the sequence.
Solution:
lim an =
n→∞
=
=
=
=
Thus, lim an = lim
n→∞
n→∞
2n
= 2.
n+1
2n
lim
n→∞
n+1
1 2n
n
lim
1
n→∞
n+1
n
2
lim
n→∞
1 + n1
2
1
since lim = 0
n→∞ n
1+0
2
(b) Determine the value of n for which an and all subsequent terms of the sequence will
1
be within 100
of the limit.
2n
is always less than 2 (since the denominan+1
of the numerator for n ∈ N). Thus, an < 2 for all values of n.
Solution: One helpful thing to note is that
tor will always be larger than
1
2
1
For the terms of the sequence to be within 100
of the limit L = 2, we need a large integer
N such that
1
|aN − L| <
100
2N
1
|
− 2| <
N +1
100
2N
1
1
<
−2<
−
100
N +1
100
2N
201
199
<
<
100
N +1
100
But as we showed above, the terms of the sequence are always less than 2 so the inequality
on the right is always satisfied. Thus, we can focus on the inequality on the left.
2N
199
<
100
N +1
199
(N + 1) < 2N
100
199
199
N+
< 2N
100
100
199
199
< 2N −
N
100
100
199
1
<
N
100
100
199 < N
1
For N > 199, every term of the sequence an is within 100
of the limit L = 2.
(c) Let be any extremely small positive number. Determine a real number N , in terms of
, such that for all natural numbers n ≥ N , an will be within of the limit.
Solution: Do not let the parameter ε deter you. The arithmetic in solving this problem is
identical to what we just did.
For the terms of the sequence to be within ε of the limit L = 2, we need a large integer
N such that
|aN − L| < ε
2N
− 1| < ε
N +1
2N
−ε <
−2<ε
N +1
2N
2−ε<
<2+ε
N +1
|
But as we showed above, the terms of the sequence are always less than 2 so the inequality
on the right is always satisfied. Thus, we can focus on the inequality on the left.
2N
N +1
(2 − ε)(N + 1) < 2N
2−ε<
(2 − ε)N + (2 − ε) < 2N
(2 − ε) < 2N − (2 − ε)N
2 − ε < 2N − 2N + εN
2 − ε < εN
2−ε
<N
ε
For N > 2−ε
, every term of the sequence an is within ε of the limit L = 2.
ε
8. Use the Limit Laws to determine the limit of the sequence generated by each function.
13n + 5n2 + 1
(a) bn :=
6 − 2n2
Solution:
13n + 5n2 + 1
lim
n→∞
6 − 2n2
1 13n + 5n2 + 1
n2
lim
1
2
n→∞
6 − 2n
n2
13
+ 5 + n12
n
lim
6
n→∞
−2
n2
13
1
lim
+ lim 5 + lim 2
n→∞ n
n→∞
n→∞ n
6
lim
− lim 2
n→∞ n2
n→∞
0+5−0
0−2
5
−
2
lim bn =
n→∞
=
=
=
=
=
(b) f (n) :=
n2 + 9
3n3 − n2 + 7n + 1
n2 + 9
lim
n→∞
3n3 − n2 + 7n + 1
1 n2 + 9
n3
lim
1
3
2
n→∞
3n − n + 7n + 1
n3
1
+ n93
n
lim
n→∞
3 − n1 + n72 + n13
1
9
lim + lim 3
n→∞ n
n→∞ n
7
1
1
lim 3 − lim + lim 2 + lim 3
n→∞ n
n→∞
n→∞ n
n→∞ n
0+0
3−0+0+0
0
lim bn =
n→∞
=
=
=
=
=
9. Let {an } and {bn } be convergent sequences where lim an = c − 1, lim bn = c2 − 1.
n→∞
n→∞
For the following sequences, first determine what is required for the new sequence to exist.
Second, assume the sequence exists and compute the limit of the sequence, if the limit exists.
Fully simplify this expression.
(a) {c2 bn − an bn }
Solution:
(i) The terms of the sequence {c2 bn − an bn } will always exist since it requires nothing more
than the multiplication and addition of numbers. This process will always be defined.
(ii)
lim (c2 bn − an bn ) =
n→∞
=
lim (c2 bn + (−1)an bn )
n→∞
lim (c2 bn ) + lim ((−1)an bn ) by limit law 1
n→∞
2
n→∞
= c lim (bn ) + (−1) lim (an bn ) by limit law 5
n→∞
n→∞
2
= c lim (bn ) − lim (an ) lim (bn ) by limit law 3
n→∞
=
=
=
=
n→∞
n→∞
c2 (c2 − 1) − (c − 1)(c2 − 1) since sequences are convergent
c4 − c2 − (c3 − c − c2 + 1)
c4 − c2 − c3 + c + c2 − 1
c4 − c3 + c − 1
n
o
bn
(b)
an +bn
Solution:
(i) The terms of the sequence exist as long as an + bn 6= 0 for any value of n.
(ii)
lim
n→∞
bn
an + b n
lim (bn )
n→∞
=
lim (an + bn )
by limit law 4
n→∞
lim (bn )
n→∞
=
lim (an ) + lim (bn )
n→∞
by limit law 1
n→∞
2
=
c −1
since sequences are convergent
(c − 1) + (c2 − 1)
For the limit to exist, we require that the denominator not be zero as well. That is,
(c2 − 1) + (c − 1) 6= 0
c2 + c − 2 6= 0
(c + 2)(c − 1) 6= 0
This requires that
c 6= −2 and c 6= 1.
Assuming c 6= −2 and c 6= 1,
lim
n→∞
bn
an + b n
c2 − 1
=
(c2 − 1) + (c − 1)
(c + 1)(c − 1)
=
(c + 2)(c − 1)
c+1
=
c+2