Math 2260 Final Exam Practice Problem Solutions
1. Write down an integral which will compute the length of the part of y = ln x for 1 ≤ x ≤ 2. Don’t
worry about evaluating this integral.
Answer: Since
dy
dx
= x1 , the arc length integral is
Z
1
2
s
2
Z 2r
1
1
1+
dx =
1 + 2 dx
x
x
1
Z 2r 2
x +1
dx
=
x2
1
√
Z 2
1 + x2
=
dx
x
1
2. Let V (a) be the volume obtained by rotating the area between the x-axis and the graph of y =
from x = a to x = 1 around the y-axis. What happens as a goes to zero?
1
x3/2
Answer: Using the cylindrical shell method, V (a) is given by
1
Z
V (a) =
2πx
dx
x3/2
a
Z
1
1
1
dx
1/2
x
a
√ 1
= 2π 2 x a
√ = 2π 2 − 2 a
√
= 4π(1 − a).
= 2π
Therefore, as a → 0, the quantity V (a) → 4π.
√
3. What is the volume of the solid obtained by rotating the part of the graph of y = cos x sin x between
x = 0 and x = π/2 around the x-axis?
Answer: Using the disk method, the volume will be
Z
π/2
Z
2
√
π cos x sin x dx =
0
π/2
π cos2 x sin x dx.
0
Then, letting u = cos x, we have that du = − sin x, and so
Z
π/2
π cos2 x sin x dx = −π
π/2
Z
0
cos2 x(− sin x) dx
0
0
Z
u2 du
= −π
1
= −π
u3
3
0
= −π 0 −
=
1
π
.
3
1
1
3
4. Find the volume obtained by rotating the region between the x-axis, the y-axis, and the line x + y = 1
around the line x = −2.
Answer: I plan to use the shell method. Since the curve is y = 1 − x, the shell based at x has radius
x − (−2) = x + 2 and height 1 − x, so the volume is
Z 1
Z 1
−x2 − x + 2 dx
2π(x + 2)(1 − x) dx = 2π
0
0
1
−x3
x2
−
+ 2x
3
2
0
−1 1
− +2
= 2π
3
2
7
= 2π
6
7π
=
.
3
√
5. Calculate the surface area obtained by rotating y = x around the x-axis for 3/4 ≤ x ≤ 15/4.
= 2π
Answer: Since
dy
dx
=
1
√
,
2 x
Z
the surface area integral is
15/4
√
s
2π x 1 +
3/4
1
√
2 x
2
Z
15/4
dx = 2π
√
r
x
1+
3/4
Z
15/4
= 2π
r
x+
3/4
1
dx
4x
1
dx.
4
Let u = x + 41 . Then du = dx and the above integral is equal to
Z
2π
4
√
u du = 2π
1
2 3/2
u
3
4
= 2π
1
16 2
28π
−
.
=
3
3
3
6. (a) Find the area between y = 3 sin(πx) and y = 6x for x > 0.
6
5
4
3
2
1
0.2
0.4
0.6
0.8
1.0
Answer: First, we need to find the points where the curves intersect. The curves intersect when
3 sin(πx) = 6x,
meaning that
sin(πx) = 2x,
which happens when x = 0 or x = 1/2. Therefore, the area of the given region will be given by
the definite integral
1/2 Z 1/2
3
−3
3
3
−3 cos(πx)
2
(3 sin(πx) − 6x) dx =
− 3x
= 0−
−
−0 = − .
π
4
π
π
4
0
0
2
(b) What is the volume obtained by rotating this region around the x-axis?
Answer: My plan is to use the washer method. The outer radius of the washer at x is 3 sin(πx),
whereas the inner radius is 6x, so the area of the washer at x is
2
A(x) = πR2 − πr2 = π (3 sin(πx)) − π(6x)2 = 9π sin2 (πx) − 36πx2 .
Therefore, the volume of the solid is
1/2
Z
1/2
Z
9π sin2 (πx) − 36πx2 dx
A(x) dx =
0
0
Z
1/2
Z
2
sin (πx) dx − 36π
= 9π
1/2
x2 dx
0
0
3 1/2
1 − cos(2πx)
x
dx − 36π
2
3 0
0
1/2
x sin(2πx)
3π
= 9π
−
−
2
4π
2
0
1
3π
= 9π
−
4
2
9π 3π
=
−
4
2
3π
.
=
4
Z
1/2
= 9π
(c) What is the volume obtained by rotating around the y-axis?
Answer: Using the shell method, when the radius of the cylindrical shell is x − 0 = x, the height
is 3 sin(πx) − 6x, so the volume of the solid is
Z
1/2
Z
1/2
2πx (3 sin(πx) − 6x) dx = 2π
0
Z
3x sin(πx) dx − 2π
0
1/2
6x2 dx.
(1)
0
1/2
R 1/2
Certainly 2π 0 6x2 = 2π 2x3 0 = π2 , which takes care of the second term. To evaluate the
first term, I will integrate by parts using
u = 3x
dv = sin(πx) dx
du = 3 dx
v=
− cos(πx)
π
Then
Z
2π
0
1/2
!
1/2 Z 1/2
−3x cos(πx)
3 cos(πx)
3x sin(πx) dx = 2π
+
dx
π
π
0
0
1/2 !
3 sin(πx)
= 2π 0 +
π2
0
=
6
.
π
Therefore, from equation (1), we have that the volume of the solid is equal to
6
π
− .
π
2
3
7. What is the volume obtained by rotating the region between y =
around the x-axis?
√ 1
x2 +4
and the x-axis for 0 ≤ x ≤ 3
Answer: Using the disk method, the radius of a cross-sectional disk is equal to
the disk at x is equal to
2
1
π
A(x) = π √
= 2
.
x +4
x2 + 4
Therefore, the volume of the solid is equal to
Z 3
Z 3
π
A(x) dx =
dx.
2+4
x
0
0
√ 1
,
x2 +4
so the area of
To evaluate this integral, make the trig substitution x = 2 tan θ. Then dx = 2 sec2 θ dθ and the above
integral becomes
Z arctan(3/2)
Z arctan(3/2)
2π sec2 θ
2π sec2 θ dθ
=
dθ
4 sec2 θ
4 tan2 θ + 4
0
0
Z arctan(3/2)
π
dθ
=
2
0
h π iarctan(3/2)
=
x
2 0
π
= arctan(3/2).
2
8. What is the length of the part of y = x3/2 from x = 1 to x = 6?
√
dy
Answer: Since dx
= 32 x1/2 = 32 x, the arc length formula is
s
2
Z 6
Z 6r
3√
9
1+
x dx =
1 + x dx.
2
4
1
1
Then we can evaluate this integral by letting u = 1 + 94 x. This yields du =
integral becomes
29/2
Z
4 29/2 √
4 2 3/2
u du =
u
9 13/4
9 3
13/4
√
√ !
8 29 29 13 13
√ −
=
27
8
2 2
√
√
58 58 − 13 13
=
.
27
R
9. Integrate x4 ln(2x) dx.
Answer: Integrate by parts with
u = ln(2x)
du =
1
dx
x
dv = x4 dx
v=
x5
.
5
Then the above integral becomes
x5 ln(2x)
−
5
Z
x4
x5 ln(2x) x5
=
−
+ C.
5
5
25
4
9
4
dx and so the above
10. Integrate
R
x2 cos(3x) dx.
Answer: Integrate by parts with
u = x2
dv = cos(3x) dx
du = 2x dx
v=
sin(3x)
3
Then the above integral is equal to
Z
Z
x2 sin(3x)
x2 sin(3x) 2
2x sin(3x)
x sin(3x) dx.
−
dx =
−
3
3
3
3
To evaluate the second integral, integrate by parts again, with
u=x
dv = sin(3x) dx
du = dx
v=
− cos(3x)
.
3
Then the above becomes
Z
x2 sin(3x) 2 −x cos(3x)
cos(3x)
x2 sin(3x) 2x cos(3x) 2 sin(3x)
−
+
=
+
−
+ C.
3
3
3
3
3
9
27
11. Integrate
R
x+2
x2 +8x+15
dx.
Answer: I plan to use partial fractions:
x2
x+2
A
B
x+2
=
=
+
+ 8x + 15
(x + 5)(x + 3)
x+5 x+3
means that
x + 2 = A(x + 3) + B(x + 5)
x + 2 = (A + B)x + (3A + 5B),
so we have
1=A+B
2 = 3A + 5B
Therefore, A = 1 − B, so the second line becomes
2 = 3(1 − B) + 5B = 3 − 3B + 5B = 3 + 2B,
and hence
2B = −1
so it follows that B = −1/2 and A = 1 − B = 3/2.
Therefore,
Z
3/2
1/2
−
dx
x+5 x+3
3
1
= ln |x + 5| − ln |x + 3| + C.
2
2
x+2
dx =
x2 + 8x + 15
Z 5
12. Integrate
R√
1 − 4x2 dx.
Answer: Make the trig substitution x =
Z r
1
2
sin θ. Then dx =
1
2
cos θ dθ and the above integral becomes
Z
1
1 p
2 1
1 − 4 · sin θ cos θ dθ =
1 − sin2 θ cos θ dθ
4
2
2
Z
1 √ 2
=
cos θ cos θ dθ
2
Z
1
cos2 θ dθ
=
2
Z
1
1 + cos(2θ)
=
dθ
2
2
1 θ sin(2θ)
=
+
+C
2 2
4
1
sin(2θ)
=
θ+
+C
4
2
2 sin θ cos θ
1
θ+
+C
=
4
2
1
= (θ + sin θ cos θ) + C,
4
using the trig identity sin(2θ) = 2 sin θ cos θ. Now, since x = 21 sin θ, it follows that θ = arcsin(2x),
√
sin θ = 2x, and cos θ = 1 − 4x2 . Therefore, converting the above back into the variable x yields
√
p
1
arcsin(2x) x 1 − 4x2
2
+
+ C.
arcsin(2x) + 2x 1 − 4x + C =
4
4
2
13. Integrate
R
sin2 (ln(x))
x
dx.
Answer: Let u = ln x. Then du =
Z
14. Integrate
R
et
e2t +4
1
x
dx, so the above becomes
sin2 (u) du =
Z
1 − cos(2u)
u sin(2u)
du = −
+ C.
2
2
4
dt.
Answer: Let u = et . Then du = et dt and the above becomes
Z
du
.
2
u +4
Now, let u = 2 tan θ. Then du = 2 sec2 θ dθ, so the above becomes
Z
Z
2 sec2 θ dθ
1
θ
=
dθ = + C.
2
2
2
4 tan θ + 4
But now, since θ = arctan(u/2) and u = et , this is just
arctan(et /2)
+ C.
2
15. Evaluate
R∞
e
1
x(ln x)2
dx.
6
Answer: Let u = ln x. Then du =
1
x
dx and the above integral becomes
∞
Z
1
1
du = lim
b→∞
u2
Z
b
u−2 du
1
b
−1
b→∞
u 1
−1
+1
= lim
b→∞
b
= lim
= 1.
16. Evaluate
R∞
0
dx
(x+1)(x+4)
dx.
Answer: Using partial fractions
1
A
B
=
+
(x + 1)(x + 4)
x+1 x+4
implies that
1 = A(x + 4) + B(x + 1)
1 = (A + B)x + (4A + B)
so
0=A+B
1 = 4A + B
Therefore B = −A and so the second equation becomes
1 = 4A − A = 3A,
so A = 1/3 and hence B = −A = −1/3.
Thus
Z
0
∞
Z
b
dx
0 (x + 1)(x + 4)
Z b
1/3
1/3
= lim
−
dx
b→∞ 0
x+1 x+4
b
1
1
ln(x + 1) − ln(x + 4)
= lim
b→∞ 3
3
0
b
1
x+1
= lim
ln
b→∞ 3
x+4 0
1
b+1
1
1
= lim
ln
− ln
b→∞ 3
b+4
3
4
1
1
= ln(1) − ln(1/4)
3
3
− ln(1/4)
=
.
3
dx
= lim
(x + 1)(x + 4) b→∞
7
17. Solve the initial-value problem
dy
+ y = 2,
dx
y(0) = 1.
Answer: My goal is to separate variables. First, notice that I can write
dy
= 2 − y.
dx
Dividing by 2 − y, multiplying by dx, and integrating yields
Z
Z
dy
= dx.
2−y
But these integrals are easy, so we get
− ln(2 − y) = x + C,
or
ln(2 − y) = −x − C.
Now I need to solve for y, so exponentiate both sides:
2 − y = e−x−C ,
or
2 − y = Ae−x
where A = e−C .
Hence
y = 2 − Ae−x .
But then, since we know y(0) = 1, we have that
y(0) = 2 − Ae−0
1 = 2 − A,
so A = 1 and the solution to the initial-value problem is
y = 2 − e−x .
18. Solve the initial-value problem
dy
= 1 − y + x2 − yx2 ,
dx
y(0) = 0.
Answer: Notice that the right hand side can be re-written as
dy
= (1 + x2 )(1 − y).
dx
Therefore, we can separate variables and integrate:
Z
Z
dy
= (1 + x2 ) dx,
1−y
so
− ln(1 − y) = x +
8
x3
+ C.
3
Now we solve for y, so multiply by − = 1 and then exponentiate both sides:
1 − y = e−x−
x3
3
−C
= e−C e−x−x
3
/3
3
= Ae−x−x
/3
where A = e−C . Hence,
3
y = 1 − Ae−x−x
/3
.
Now we can use the initial condition to solve for A:
y(0) = 1 − Ae−0−0
3
/3
0 = 1 − A,
so A = 1 and therefore the solution to the initial-value problem is
3
y = 1 − e−x−x
/3
.
∞
19. Consider the sequence (an )n=1 , where
an =
ln(1 + 3/n)
.
sin(2/n)
Does the sequence converge? If so, to what? If not, explain why.
Answer: Taking the limit yields
ln(1 + 3/n)
n→∞ sin(2/n)
ln(1 + 3/x)
= lim
x→∞ sin(2/x)
lim an = lim
n→∞
=
1
−3
1+3/x x2
lim
x→∞ cos(2/x) −2
x2
= lim
x→∞
=
3
2 cos(2/x) 1 + x3
3
,
2
where I got from the second to the third lines by L’Hôpital’s Rule, so the sequence converges to 32 .
∞
20. Consider the sequence (an )n=1 , where
an =
n+1
n+2
n
.
Does the sequence converge? If so, to what? If not, explain why.
Answer: The goal is to evaluate limn→∞ ln(an ); then we’ll have that
lim an = elimn→∞ ln(an ) .
n→∞
9
Now,
n n+1
n→∞
n+2
n+1
= lim n ln
n→∞
n+2
x+1
ln x+2
= lim
1
lim ln(an ) = lim ln
n→∞
x→∞
= lim
x
ln(x + 1) − ln(x + 2)
1
x
x→∞
= lim
x→∞
1
x+1
−
1
x+2
−1
x2
2
−x
x2
+
x→∞ x + 1
x+2
2
−x (x + 2) + x2 (x + 1)
= lim
x→∞
(x + 1)(x + 2)
−x2
= lim
x→∞ (x + 1)(x + 2)
−x2
= lim 2
x→∞ x + 3x + 2
= −1,
= lim
where I went from the fourth to the fifth lines using L’Hôpital’s Rule.
Therefore,
lim an = e−1 =
n→∞
1
.
e
21. Which of the following series converge?
P∞ n3
(a)
n=1 3n
Answer: Using the Ratio Test:
(n+1)3
n+1
lim 3 n3
n→∞
3n
(n + 1)3 3n
n→∞ 3n+1 n3
= lim
1 n3
n→∞ 3 (n + 1)3
1
= .
3
= lim
(b)
Since
P∞
1
3
< 1, the Ratio Test says that this series converges.
n3
n=1 1+n4
Answer: I will do a limit comparison to the series
n3
1+n4
3
n→∞ n
n4
lim
P n3
n4 :
n4
= 1.
n→∞ 1 + n4
= lim
P n3
P1
Since the series
n4 =
n diverges and the above limit is finite and positive, the Limit Comparison Test implies that the given series also diverges.
10
(c)
n3
n=1 1+n5
P∞
Answer: Since 1 + n5 > n5 , it follows that
n3
1
n3
<
= 2
5
5
1+n
n
n
for all n, and so
(d)
∞
X
∞
X
n3
1
<
.
5
2
1
+
n
n
n=1
n=1
Since the series on the right hand side converges, so does the given series.
P∞ cos2 n
n=1 1+e−n
Answer: Notice that
cos2 n
cos2 n
= lim
,
−n
n→∞
n→∞ 1 + e
1
which does not exist. Therefore, since the terms are not going to zero, the nth term test implies
that the series diverges.
P ∞ en
lim
(e)
n=1 n!
Answer: Using the Ratio Test:
en+1
(n+1)!
lim
en
n→∞
n!
en+1 n!
n→∞ (n + 1)! en
e
= lim
n→∞ n + 1
= 0.
= lim
Since 0 < 1, the Ratio Test implies that the series converges.
22. If you put (−1)n into each of the series in the preceding problem, which ones converge absolutely,
which converge conditionally, and which diverge?
Answer:
(a) The answer to the previous problem shows that series
P
3
(−1)n 3nn converges absolutely.
3
n
(b) The series cannot converge absolutely, but since the terms 1+n
4 satisfy the hypotheses of the
P
3
n n
Alternating Series Test, the series (−1) 1+n4 does converge, so we conclude that it converges
conditionally.
P
n3
(c) The answer to the previous problem shows that the series (−1)n 1+n
5 converges absolutely.
P
2
cos n
(d) The terms still don’t go to zero, so the series (−1)n 1+e
−n diverges.
P
n
(e) The answer to the previous problem shows that the series (−1)n en! converges absolutely.
23. Find the interval of convergence for each of the following power series.
P∞ (x+2)n
(a)
n=1
n
Answer: Using the Ratio Test on the series of absolute values,
(x+2)n+1 n+1 |x + 2|n+1
n
= lim
lim n
(x+2)
n→∞
n→∞
n + 1 |x + 2|n
n n
= lim |x + 2|
n→∞
n+1
= |x + 2|.
11
Therefore, the series definitely converges absolutely when |x + 2| < 1.
Now check the endpoints. When x + 2 = 1, the series becomes
∞
∞
X
X
1n
1
=
,
n
n
n=1
n=1
which diverges.
When x + 2 = −1, the series becomes
∞
X
(−1)n
,
n
n=1
which converges. Therefore, the power series converges for
−1 ≤ x + 2 < 1,
or
−3 ≤ x < −1,
(b)
which means the interval of convergence is [−3, −1).
P∞ (x−3)n
n=1 1+n2
Answer: Using the Ratio Test on the series of absolute values,
(x−3)n+1 1+(n+1)2 |x − 3|n+1 1 + n2
= lim
lim n
(x−3) n→∞
n→∞ 1 + (n + 1)2 |x − 3|n
1+n2 1 + n2
n→∞
1 + (n + 1)2
n2 + 1
= lim |x − 3| 2
n→∞
n + 2n + 2
= |x − 3|.
= lim |x − 3|
Therefore, the series definitely converges absolutely when |x − 3| < 1.
Now check the endpoints. When x − 3 = 1, the series becomes
∞
∞
X
X
1
1
1n
=
<
,
2
2
2
1
+
n
1
+
n
n
n=1
n=1
n=1
∞
X
which converges, so the series converges when x − 3 = 1.
When x − 3 = −1, the series becomes
∞
X
(−1)n
,
1 + n2
n=1
which converges by the Alternating Series Test.
Therefore, the given power series converges for
−1 ≤ x − 3 ≤ 1,
or
2 ≤ x ≤ 4,
so the interval of convergence is [2, 4].
12
(c)
P∞
n=1
(−1)n (x−1)n
nen
Answer: Using the Ratio Test on the series of absolute values,
(−1)n+1 (x−1)n+1 (n+1)en+1 nen
|x − 1|n+1
lim =
lim
(−1)n (x−1)n n→∞
n→∞ (n + 1)en+1 |x − 1|n
nen
|x − 1| n
e n+1
|x − 1|
=
.
e
= lim
n→∞
< 1, meaning for |x − 1| < e.
Therefore, the series definitely converges absolutely for |x−1|
e
Now, check the endpoints. When x − 1 = e, the series becomes
∞
∞
X
X
(−1)n en
(−1)n
=
,
n
ne
n
n=1
n=1
which converges.
When x − 1 = −e, the series becomes
∞
∞
∞
X
X
X
(−1)n (−e)n
(−1)n (−1)n en
1
=
=
,
n
n
ne
ne
n
n=1
n=1
n=1
which diverges.
Therefore, the given power series converges for
−e < x − 1 ≤ e,
or
1 − e < x ≤ 1 + e,
(d)
so the interval of convergence is (1 − e, 1 + e].
P∞
2 n
n=1 n x
Answer: Using the Ratio Test on the series of absolute values,
(n + 1)2 xn+1 (n + 1)2 |x|n+1
=
lim
lim
n→∞
n→∞
|n2 xn |
n2
|x|n
2
(n + 1)
= lim
|x|
n→∞
n2
= |x|,
so the series definitely converges absolutely for |x| < 1.
Now check the endpoints. When x = 1, the series becomes
∞
X
n2 1n =
n=1
∞
X
n2 ,
n=1
which diverges since the terms are going to infinity.
Likewise, when x = −1, the series becomes
∞
X
n=1
13
n2 (−1)n ,
which diverges since the terms are getting large in absolute value and hence not going to zero.
Therefore, the series converges for
−1 < x < 1,
so the interval of convergence is (−1, 1).
24. Find the Taylor series for
2
(a) e−x centered at x = 0
Answer: The Taylor series for ex centered at x = 0 is
∞
X
x3
x2
xn
+
+ ... =
.
e =1+x+
2!
3!
n!
n=0
x
2
Then, substituting −x2 for x in the above yields the Taylor series for e−x centered at x = 0:
2
(−x2 )2
(−x2 )3
+
+ ...
2!
3!
4
6
x
x
= 1 − x2 +
−
+ ...
2!
3!
∞
X
(−1)n x2n
=
.
n!
n=0
e−x = 1 + (−x2 ) +
(b)
√
x centered at x = 1
Answer: Remember that the Taylor series for a function f (x) centered at x = c is given by
∞
X
f (n) (c)
f (x) =
(x − c)n .
n!
n=0
Therefore, we need to determine the values of the derivatives of f (x) =
compute the derivatives of f (x):
1
f 0 (x) = √
2 x
−1
00
f (x) = 3/2
4x
3
f 000 (x) = 5/2
8x
−15
f (4) (x) =
16x7/2
..
.
Therefore, values of f and its derivatives at x = 1 are:
√
f (1) = 1 = 1
1
1
f 0 (1) = √ =
2
2 1
−1
−1
f 00 (1) =
=
4
4 · 13/2
3
3
f 000 (1) =
=
8
8 · 15/2
−15
−15
f (4) (1) =
=
16
16 · 17/2
..
.
14
√
x at x = 1. First,
Therefore, the Taylor series for f is
1
1
3
15
f (x) = 1 + (x − 1) −
(x − 1)2 +
(x − 2)3 −
(x − 1)4 + . . .
2
4 · 2!
8 · 3!
16 · 4!
25. Estimate e−0.2 to the nearest 0.001. Explain your answer.
Answer: Since the Taylor series centered at x = 0 for ex is
ex = 1 + x +
x2
x3
+
+ ...,
2!
3!
and this series converges absolutely for all real numbers x, it follows that
e−0.2 = 1 + (−0.2) +
(−0.2)2
(−0.2)3
0.04 0.008
+
+ . . . = 1 − 0.2 +
−
+ ....
2!
3!
2
6
4
= 0.0016
< 0.001, it follows that the
Since the series alternates and the next term in the series is (0.2)
4!
24
approximation
0.04 0.008
−
= 0.8 + 0.02 − 0.0013 ≈ 0.819
e−0.2 ≈ 1 − 0.2 +
2
6
is correct to three decimal places.
26. Does the series
∞
X
n7 6n
n!
n=1
converge or diverge? Explain your answer.
Answer: Use the Ratio Test:
(n+1)7 6n+1
(n+1)!
lim
n7 6n
n→∞
n!
(n + 1)7 6n+1
n!
6
(n + 1)7
· 7 n = lim
·
=0
n→∞
n→∞ n + 1
(n + 1)!
n 6
n7
= lim
since the first term goes to zero and the second term goes to 1.
Therefore, since 0 < 1, the Ratio Test tells us that the series converges.
27. Does the series
∞
X
1
√
n
+
n
n=2
converge or diverge? Explain your answer.
Answer:
P 1 For large n, the n should dominate the
series
n.
1√
n+ n
1
n→∞
n
lim
Therefore, since
P
1
n
√
n, so let’s do a limit comparison to the divergent
n
1
n
1
√ · = lim
√ = lim
= 1.
n→∞ n +
n+ n 1
n n→∞ 1 + √1n
P 1
√ .
diverges, so does
n+ n
= lim
n→∞
28. Find the limit
lim
x→0
cos x − 1
ex2 − 1
using Taylor series.
Answer: Recall that the Taylor series centered at x = 0 for cos(x) is
cos(x) = 1 −
x2
x4
x6
+
−
+ ....
2!
4!
6!
15
Since ex = 1 + x +
x2
2!
+
x3
3!
2
+ . . ., the Taylor series centered at x = 0 for ex is
2
ex = 1 + (x2 ) +
(x2 )2
(x2 )3
x4
x6
+
+ . . . = 1 + x2 +
+
+ ....
2!
3!
2!
3!
Therefore,
2
1 − x2! +
cos(x) − 1
= lim
lim
2
x→0 1 + x2 +
x→0 ex − 1
2
= lim
x→0
= lim
x4
2!
− ... − 1
+ ... − 1
x4
24 − . . .
x4
2 + ...
− x2 +
x2 +
x2 − 12 +
x2 1 +
x→0
= lim
x4
4!
− 12 +
1+
x→0
x4
24
x2
2
− ...
+ ...
x2
24
x2
2
− ...
+ ...
1
=− .
2
29. Give a geometric description of the set of points (x, y, z) satisfying the following inequalities.
(a) x2 + y 2 + z 2 ≤ 1 and z ≥ 1/2.
Answer: Since x2 + y 2 + z 2 = 1 describes the surface of the sphere of radius 1, the points
satisfying the inequality x2 + y 2 + z 2 ≤ 1 are those points lying on the surface of this sphere or
inside it.
On the other hand, the plane z = 1/2 is the plane parallel to the xy-plane at height 1/2; the
points satisfying z ≥ 1/2 are those points lying in this plane or above it.
Therefore, the points satisfying both of the given inequalities are those points lying on or in the
sphere of radius 1 centered at the origin and above the plane at height 1/2; in other words, this
is the top one-fourth of the sphere of radius 1.
(b) |x| ≤ 1, |y| ≤ 1, and |z| ≤ 1.
Answer: If |x| ≤ 1, then −1 ≤ x ≤ 1, so points satisfying this inequality are those points lying
no further than a distance 1 from the yz-plane, and similarly for the other two inequalities.
Therefore, the points satisfying all of the above inequalities are those points lying within the cube
of side length 2 which is centered at the origin.
30. Determine the angle between the vectors
*
*
*
*
u = ı + 3  + 2k
*
*
*
*
v = −2 ı +  − 2 k
and
Answer: Recall that
*
*
*
*
u · v = | u|| v| cos θ,
*
*
where θ is the angle between u and v, so it follows that
*
cos θ =
*
u·v
* * .
| u|| v|
Now, by definition of the dot product,
*
*
u · v = (1)(−2) + (3)(1) + (2)(−2) = −2 + 3 − 4 = −3.
16
Also,
*
| u| =
p
12 + 32 + 22 =
√
and
*
| v| =
p
(−2)2 + 12 + (−2)2 =
1+9+4=
√
4+1+4=
Therefore,
*
cos θ =
*
*
u·v
−3
−1
√ =√ ,
* * =
| u|| v|
3 14
14
*
so the angle between u and v is
θ = arccos
(If you’re curious, this turns out to be about 105◦ .)
17
√
−1
√
14
.
14
√
9 = 3.