MATH 166 (F-2, G-2, J-2)
Practice Questions, Thu. Dec. 12
1. Find the limit: lim
x→0
x2 sin (1/x)
tan x
Solution:
We know −1 ≤ sin θ ≤ 1 for any value of θ. So, −1 ≤ sin (1/x) ≤ 1, which is the same as
| sin (1/x)| ≤ 1 Therefore,
2
2
0/0f orm,L0 Hopital
x sin (1/x) 2x
0
≤ lim x
≤
= lim
= =0
lim 2
x→0 tan x
x→0 sec x
x→0
tan x
1
2
x sin (1/x)
Therefore, by the squeeze theorem, lim
= 0.
x→0
tan x
2. Evaluate the following improper integrals
Z ∞
(a)
xe−2x dx
0
Solution: We use integration by parts; use u = x and dv = e−2x dx
Z ∞
xe−2x dx
0
Z ∞ −2x
xe−2x ∞
e
dx
=
−
−2 0
−2
0
xe−2x ∞
e−2x ∞
=
−
−2 0
(−2)(−2) 0
xe−2x ∞ e−2x ∞
=
−
−2 0
4 0
ae−2a (0)e−2(0)
e−2b (e−2(0)
= lim
−
− lim
−
a→∞ −2
b→∞ 4
−2
4
−2a
ae
0 1
= lim
−0 −
−
a→∞ −2
4 4
ae−2a 1
= lim
+
a→∞ −2
4
This limit is of the form (∞.0), which is an indeterminate form, but, we can’t apply L’Hopital’s
rule yet. Rewriting the exponential function in the denominator:
a
1
+
2a
a→∞ −2e
4
= lim
Now, this is of the form ∞/∞, for which we can apply L’Hopital’s rule. Applying L’Hopital’s
rule, we get:
= lim
a→∞
1
1
1
1
+ =0+ = .
2a
(−2)(2)e
4
4
4
Z
∞
(b)
−∞
x2
1
dx
+ 2x + 10
Solution: Completing the square:
Z ∞
Z ∞
Z ∞
1
1
1
dx =
dx =
dx
2
2
2
2
−∞ (x + 1) + 9
−∞ (x + 1) + 3
−∞ x + 2x + 10
Z 0
Z b
1
1
= lim
dx
+
lim
dx
a→∞ −a (x + 1)2 + 32
b→∞ 0 (x + 1)2 + 32
0
b
x + 1 x + 1 1
1
−1
−1
lim tan
= lim tan
+ b→∞
a→∞ 3
3
3
3
−a
0
1
−a
+
1
b+1
1
1
−1 −1
−1
−1 =
tan − lim tan
+ lim tan
− tan a→∞
b→∞
3 3
3
3
3
1
−π π
=
−
+
3
2
2
π
=
3
3. Determine if each of the following series converges or diverges:
∞
X
n2 + n + 10
(a)
4n2 + 5
n=1
n2 + n + 10
1
Solution: Note that lim
= 6= 0
2
n→∞
4n + 5
4
Hence, by the nth term divergence test, the series diverges.
∞
X
(−1)n
√
(b)
2n
n
n=1
Solution:
1
1
√ = lim n− 2n is of the form ∞0
Consider lim 2n
n→∞
n n→∞
1
1
ln n
Let y = n− 2n . Then, ln y = − ln n = −
.
2n
2n
As n → ∞, this is of the form ∞/∞, for which we can apply L’Hopital’s rule. So,
lim ln y = lim −
n→∞
n→∞
ln n
1/n
= lim −
=0
n→∞
2n
2
1
Hence, lim y = lim n− 2n = e0 = 1 6= 0. Therefore, the series must diverge.
n→∞
n→∞
Page 2
(c)
∞
X
2n2 + n + 7
3n3 + 2n + 1
n=1
Solution: We use Limit Comparison test.
1
Pick bn = . Then,
n
2n2 +n+7
(2n3 + n2 + 7n)
2
an
(2n2 + n + 7)n
3
= lim 3n +2n+1
=
lim
= >0
lim
=
lim
1
3
3
n→∞
n→∞ (3n + 2n + 1)
n→∞ bn
n→∞ (3n + 2n + 1)
3
n
∞
∞
X
X 1
We know
bn =
diverges since it is the Harmonic Series.
n
n=1
n=1
∞
∞
X
X
2n2 + n + 7
Therefore,
an =
also diverges by the Limit Comparison test.
3 + 2n + 1
3n
n=1
n=1
(d)
∞
X
1
n=2
n(ln n) 2
5
Solution: We use the Integral test.
Z ∞
1
1
Let u = ln x, Then du =
dx
5 dx
x
x(ln x) 2
2
Z
x=∞
=
x=2
Z
1
(u)
5
2
x=∞
du =
5
u− 2 du
x=2
x=∞
∞
3
u− 2 −2 =
=
−3/2 3(ln x)3/2 2
x=2
−2
−2
−
3/2
a→∞ 3(ln a)
3(ln 2)3/2
−2
=0−
3(ln 2)3/2
2
=
, Which is a finite number.
3(ln 2)3/2
∞
X
1
Therefore, by the integral test,
5 also converges.
2
n=2 n(ln n)
= lim
Page 3
(e)
∞
X
1
1
√ −√
n
n+1
n=1
Solution: This is a collapsing series.
N
∞
X
X
1
1
1
1
√ −√
√ −√
= lim
We know by definition
n
n + 1 N →∞ n=1 n
n+1
n=1
N
X 1
1
√ −√
Consider
,
n
n+1
n=1
N
X
1
1
√ −√
n
n+1
n=1
! 1
1
1
1
1
1
1
1
1
1 + √ −√
= √ − √ + √ − √ + √ − √ +· · ·+ √ − √
N
−
1
N +1
1 2
N
N
2
3
3
4
1
1
=√ −√
N +1
1
∞
N
X
X
1
1
1
1
1
1
1
√ −√
√ −√
So,
= lim
= lim √ − √
= √ −0=1
n
n + 1 N →∞ n=1 n
n + 1 N →∞ 1
N +1
1
n=1
4. Find the convergence set for the power series:
∞
X
(2n − 1)xn
n=1
n!
an+1 (2(n + 1) − 1)xn+1
n!
= lim Solution: lim ·
n→∞
an n→∞ (n + 1)!
(2n − 1)xn (2n + 1)
= |x| · 0 = 0 < 1 for all values of x.
= |x| lim
n→∞ (2n − 1)(n + 1)
Therefore, the series converges for all values of x, so, the radius of convergence is ∞, interval of
convergence is: (−∞, ∞)
Page 4
5. Find the radius of convergence for the power series
∞
X
n=1
n2 3n
(x − 1)n
(2n + 5n )
an+1 (n + 1)2 3n+1 (x − 1)n+1
(2n + 5n ) Solution: lim · 2 n
= lim
n→∞
an n→∞ (2n+1 + 5n+1 )
n 3 (x − 1)n 0
:
n
(2/5)
+ (5/5)n )
(n + 1)2
(2n + 5n )
=
3
|x
−
1|
·
1
·
lim
= 3 |x − 1| lim
·
lim
:0
n→∞ (2 (2/5)
n→∞
n→∞ (2n+1 + 5n+1 )
n
n2
+ 5(5/5)n+1
3
1
= 3 |x − 1| = |x − 1|
5
5
3
For convergence, we need |x − 1| < 1
5
5
=⇒ |x − 1| <
3
8
2
=⇒ − < x <
3
3
5
So, the radius of convergence is: .
3
Page 5
6. Find a power series for the function f (x) =
Solution: We know
1
.
(3 − 2x)2
1
= 1 + t + t2 + t3 + · · · + tn + · · ·
1−t
for |t| < 1
1
1
1
Now,
=
3 − 2x
3 1 − 2x
3
"
#
2 3
n
2x
2x
2x
2x
1
1+
+
+
+ ··· +
+ ···
for
=
3
3
3
3
3
"
#
2
3
n
1 1 2x
1 2x
1 2x
1 2x
=
+
+
+
+ ··· +
+ ···
3 3 3
3 3
3 3
3 3
1 2x 22 x2 23 x3
2n xn
=
+
+ 3 + 4 + · · · + n+1 + · · ·
3 32
3
3
3
2x <1
3
Differentiating both sides with respect to x (term-by-term differentiation of the right hand side)
2
22 2x 23 3x2
2n nxn−1
(−1)(−2)
= 0+ 2 + 3 +
+ ··· +
+ ···
(3 − 2x)2
3
3
34
3n+1
=⇒
2
2
22 2x 23 3x2
2n nxn−1
=
+
+
+
·
·
·
+
+ ···
(3 − 2x)2
32
33
34
3n+1
=⇒
1
2
22 2x 23 3x2
2n nxn−1
=
+
+
+
·
·
·
+
+ ···
(3 − 2x)2
2 · 32 2 · 33
2 · 34
2 · 3n+1
=
1
2 · 2x 22 · 3x2
2n−1 nxn−1
+
+
+
·
·
·
+
+ ···
32
33
34
3n+1
Page 6
for |x| <
3
2
1
, find the Taylor polynomial of order 2, P2 (x), based at 1. Estimate the absolute
x−3
value of the error R2 (x) for 0 ≤ x ≤ 2.
7. For f (x) =
Solution: We are given that a = 1.
1
x−3
−1
f 0 (x) =
(x − 3)2
2
f 00 (x) =
(x − 3)3
−6
f (3) (x) =
(x − 3)4
1
2
1
f 0 (1) = −
4
2
f 00 (1) = −
8
f (1) = −
f (x) =
f (3) (1) = −
6
16
So,
P2 (x) = f (1) +
f 0 (1)(x − 1) f 00 (1)(x − 1)2
+
1!
2!
1 − 1 (x − 1) − 28 (x − 1)2
=− + 4
+
2
1!
2!
f (3) (c)(x − 1)3
=
We know R2 (x) =
3!
−6
(x
(c−3)4
− 1)3
3!
=
−(x − 1)3
(c − 3)4
For 0 ≤ x ≤ 2, |R2 (x)| is maximum when c = 2 and x = 2 (maximum numerator and minimum denominator).
−(2 − 1)3 =1
So, |R2 (x)| ≤ (2 − 3)4 Page 7
8. Find the area of the region inside the cardioid r = 2 + 2 cos θ and outside the circle r = 3.
Solution:
90◦
60◦
180◦
0◦
300◦
◦
270
To find the intersection points, set 2 + 2 cos θ = 3 =⇒ cos θ =
Z
π
3
So, the area is:
1
2
Z
1
=
2
Z
1
=
2
Z
=
1
=
2
π
3
− π3
1
1
(2 + 2 cos θ)2 − (3)2 dθ
2
2
8 cos θ + 4 cos2 θ − 5 dθ
− π3
π
3
8 cos θ +
− π3
π
3
4(1 + cos 2θ)
− 5 dθ
2
8 cos θ + 2 cos 2θ − 3 dθ
− π3
√
π3
2 sin 2θ
9
3
8 sin θ +
− 3θ =
−π
2
2
−π
3
Page 8
π
1
=⇒ θ = ± .
2
3
9. Find the area of the region in the first quadrant, inside the two cardioids r = 2 + 2 sin θ and r =
2 + 2 cos θ.
Solution:
90◦
45◦
180◦
0◦
270◦
π 5π
To find the intersection points, set 2 + 2 sin θ = 2 + 2 cos θ =⇒ tan θ = 1 =⇒ θ = , .
4 4
Z π
Z π
4 1
2 1
So, the area is:
(2 + 2 sin θ)2 dθ +
(2 + 2 cos θ)2 dθ
π
2
2
0
Z π
Z π4
4
1
1 2
2
4 + 8 cos θ + 4 cos2 θ dθ
=
4 + 8 sin θ + 4 sin θ dθ +
2 0
2 π
4
Z π
Z π
1 4
4(1 − cos 2θ)
1 2
4(1 + cos 2θ)
=
4 + 8 sin θ +
) dθ +
4 + 8 cos θ +
dθ
2 0
2
2 π4
2
√
3π
.
=7−4 2+
2
Page 9