Math 115 HW #2 Solutions
1. In the special theory of relativity, the mass of a particle with velocity v is given by
m0
m= p
1 − v 2 /c2
where m0 is the mass of the particle at rest and c is the speed of light. What happens as
v → c− ?
Answer: As v → c− , the fraction
v2
c2
gets closer and closer to 1 (though it is always less than 1 since v is approaching c from the
left). Hence, the quantity
v2
1− 2
c
−
is approaching zero. Therefore, as v → c , the mass:
m= p
2. Evaluate
m0
1 − v 2 /c2
lim
t→0
→ +∞.
1
1
− 2
t
t +t
.
Answer: Notice that t2 + t = t(t + 1). Therefore,
1
1
t+1
1
t
1
− 2
=
−
=
=
.
t
t +t
t(t + 1) t(t + 1)
t(t + 1)
t+1
Therefore,
lim
t→0
1
1
− 2
t
t +t
3. Evaluate
lim √
x→∞
1
= 1.
t→0 t + 1
= lim
x
.
+1
x2
Answer: Using L’Hôpital’s Rule,
x
lim √
= lim
x→∞
x2 + 1 x→∞
1
1√ 1
2 x2 +1
· 2x
= lim
1
x→∞ √ x
x2 +1
√
= lim
x→∞
x2 + 1
,
x
provided that limit exists. If we knew the limit existed, then it would have to be 1, since the
left and right sides are reciprocals of each other. However, we don’t know, a priori, that the
limit exists, so we must use another method.
Let’s divide both the numerator and denominator by x:
lim √
x→∞
1
x
x
1
1
= lim 1 √ x
= lim p
= lim p
= 1.
2
2
2
2
2
x→∞
x + 1 x→∞ x x + 1 x→∞ x /x + 1/x
1 + 1/x2
1
Stewart §12.1
22. Let
3n+2
.
5n
Determine whether the sequence (an ) converges or diverges. If it converges, find the limit.
an =
Answer: We can re-write the terms in the sequence as
n
3n+2
3
an = n = 9 ·
.
5
5
n
Since 35 → 0 as n → ∞, we see that
n
3n+2
3
lim
= 9 lim
= 0.
n
n→∞ 5
n→∞ 5
28. Let
an = cos(2/n).
Determine whether the sequence (an ) converges or diverges. If it converges, find the limit.
Answer: Consider the function f (x) = cos(2/x). Then
2
lim f (x) = lim cos(2/x) = cos lim
= cos(0) = 1.
x→∞
x→∞
x→∞ x
Therefore,
lim cos(2/n) = lim f (x) = 1.
n→∞
x→∞
34. Let
an = n cos nπ.
Determine whether the sequence (an ) converges or diverges. If it converges, find the limit.
Answer: Notice that cos nπ = (−1)n , so we can re-write the terms as
an = n cos nπ = n(−1)n .
The sequence is unbounded, so it diverges.
42. Let
(ln n)2
n
Determine whether the sequence (an ) converges or diverges. If it converges, find the limit.
an =
Answer: Consider the function f (x) =
(ln x)2
x .
Then, using L’Hôpital’s Rule,
(ln x)2
x→∞
x
2 ln x x1
= lim
x→∞
1
2 ln x
= lim
.
x→∞
x
lim f (x) = lim
x→∞
2
Using L’Hôpital’s Rule again, this limit is equal to
2 x1
2
= lim
= 0.
x→∞ 1
x→∞ x
lim
Therefore,
(ln x)2
(ln n)2
= lim
= 0.
x→∞
n→∞
n
x
lim
54. (a) Determine whether the sequence defined as follows is convergent or divergent:
a1 = 1 an+1 = 4 − an
for n ≥ 1.
Answer: Writing down the first few terms of the sequence, we see that
a1 = 1,
a2 = 3,
a3 = 1,
a4 = 3,
a5 = 1, . . .
The terms oscillate between 1 and 3, so the sequence cannot converge.
(b) What happens if the first term is a1 = 2?
Answer: If a1 = 2, then the sequence looks like
a1 = 2,
a2 = 2,
a3 = 2,
a4 = 2,
a5 = 2, . . .
which is as convergent as a sequence could possibly be.
58(b). A sequence (an ) is defined by a1 = 1 and an+1 = 1/(1 + an ) for n ≥ 1. Assuming that (an )
is convergent, find its limit.
Answer: Suppose limn→∞ an = L. Then we have that
L = lim an = lim an+1
n→∞
by shifting the index
n→∞
= lim
n→∞
=
1
1 + an
by definition of an+1
1
1+L
by taking the limit.
Therefore,
1
,
1+L
meaning that, if we multiply both sides by 1 + L,
L=
1 = L(1 + L) = L + L2 .
Equivalently,
0 = L2 + L − 1,
√
so L =
−1± 5
.
2
But since all the terms in the sequence are positive, L cannot be negative, so
√
−1 + 5
L=
.
2
3
60. Determine whether the sequence
an = (−2)n+1
is increasing, decreasing, or not monotonic. Is the sequence bounded?
Answer: The first few terms of the sequence are
−2, 4, −8, 16, −32, . . .
The sequence is neither monotonic nor bounded.
62. Determine whether the sequence
2n − 3
3n + 4
is increasing, decreasing, or not monotonic. Is the sequence bounded?
an =
Answer: Let f (x) =
f 0 (x) =
2x−3
3x+4 .
Then
6x + 8 − (6x − 9)
17
2(3x + 4) − 3(2x − 3)
=
=
> 0,
(3x + 4)2
(3x + 4)2
(3x + 4)2
so f (x) is an increasing function. Therefore, since f (n) = an , the sequence (an ) is increasing
as well.
Since the sequence is increasing, each term an > a1 =
an =
−1
7 .
On the other hand, each term
2n − 3
2n − 3
2n
2
<
<
= .
3n + 4
3n
3n
3
Therefore, the terms are trapped:
2
−1
< an <
7
3
for all n,
so the sequence is bounded.
70. Show that the sequence defined by
a1 = 2 an+1 =
1
3 − an
satisfies 0 < an ≤ 2 and is decreasing. Deduce that the sequence is convergent and find its
limit.
Answer: First, we prove by induction that 0 < an ≤ 2 for all n.
0: Clearly, 0 < a1 ≤ 2 since a1 = 2.
1: Assume 0 < an ≤ 2.
2: Then, using that assumption,
an+1 =
1
1
1
>
= > 0.
3 − an
3−0
3
Also,
1
1
≤
= 1 < 2,
3 − an
3−2
so we see that the assumption that 0 < an ≤ 2 implies that
an+1 =
0 < an+1 < 2.
4
3: Therefore, by induction, 0 < an ≤ 2 for all n.
Next, we show that an+1 ≤ an for all n, again using induction.
0: Clearly, a2 ≤ a1 since a2 = 1 and a1 = 2.
1: Assume an+1 ≤ an .
2: Then, using that assumption,
an+2 =
1
1
≤
= an+1 .
3 − an+1
3 − an
3: Therefore, by induction, an+1 ≤ an for all n.
We’ve shown that the sequence (an ) is bounded and decreasing, so the Monotone Convergence
Property implies that it converges. Call the limit of the sequence L. Then
L = lim an = lim an+1
n→∞
by shifting the index
n→∞
1
n→∞ 3 − an
1
=
3−L
= lim
by definition of an+1
by taking the limit.
Therefore,
L(3 − L) = 1,
or, equivalently,
L2 − 3L + 1 = 0.
√
Possible solutions to this equation are L = 3±2 5 , but, since all the terms of the sequence are
between 0 and 2, L must be between 0 and 2 as well. Therefore,
√
3− 5
L=
.
2
5