MA 1001B Calculus: Homework Assignment 8
Class:
Score:
Student Number:
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Due 2013/12/26
(1). Find the limit: lim
x + 2 x
x→∞
x−1
. [§7.5, p403, #61]
Solution:
x + 2 x
=⇒ lim f (x) = lim eln f (x) = elimx→∞ ln f (x)
x→∞
x→∞
x + 2
ln x+2
x−1
0
∵ lim ln f (x) = lim x ln
=(∞·0) lim
=( 0 ) lim
1
x→∞
x→∞
x→∞
x→∞
x−1
x
Let f (x) =
x−1
1
x+2
−
1
x−1
−1
x2
3x2
3x2
3
= lim 2
= lim
=3
−1
x→∞ (x + 2)(x − 1)
x→∞ x + x − 2
x→∞ 1 + x
− 2x−2
x + 2 x
∴ lim
= e3 .
x→∞ x − 1
tan(2x)
a
sin(bx) (2). For what values of a and b is lim
+
+
= 0? [§7.5, p403, #80]
x→0
x3
x2
x
Solution:
tan(2x)
0
a
sin(bx) tan(2x) + ax + x2 sin(bx)
∵ lim
+
+
=
lim
x→0
x→0
x3
x2
x
x3
0
2
2
2 sec (2x) + a + 2x sin(bx) + bx cos(bx)
0
= lim
2
x→0
3x
0
2
2
∴ lim 2 sec (2x) + a + 2x sin(bx) + bx cos(bx) = 0 =⇒ a = −2
= lim
x→0
0
2 sec2 (2x) − 2 + 2x sin(bx) + bx2 cos(bx)
x→0
3x2
0
0
2
8 sec (2x) tan(2x) + 2 sin(bx) + 4bx cos(bx) − (bx)2 sin(bx)
= lim
x→0
6x
0
2
2
4
2
3 2
32 sec (2x) tan (2x) + 16 sec (2x) + 6b cos(bx) − 6b x sin(bx) − b x cos(bx)
= lim
x→0
6
0 + 16 + 6b − 0 − 0
16 + 6b
8
=
=
= 0 =⇒ b = −
6
6
3
−1/x2
e
,
x 6= 0,
(3). Find f 0 (0) for f (x) =
[§7.5, p403, #88]
0,
x = 0.
∴ lim
Solution:
2
1
f (0 + h) − f (0)
e−1/h
h
= lim
= lim 1/h
2
x→0
x→0
x→0 e
h
h
−1
1
h
1
2
2
= lim −2 h1/h2 = lim 1/h2 = lim he−1/h = 0
x→0
x→0
x→0 3 e
2
2
e
h
f 0 (0) = lim
1
∞
∞
Z
(4). Evaluate the integral:
dx
p
. [§7.6, p413, #59]
(2x − 1) (2x − 1)2 − 4
Solution:
du
Let u = 2x − 1 =⇒
=2
dx
Z
Z
dx
du
1
√
p
∴
=
2
u u2 − 4
(2x − 1) (2x − 1)2 − 4
du
Let u = 2 sec(θ) =⇒
= 2 sec(θ) tan(θ)
dθ
Z
Z
Z
du
2 sec(θ) tan(θ)
1
1
1
√
dθ =
∴
=
1 dθ
2
2
2 sec(θ)2 tan(θ)
4
u u2 − 4
θ
1
1
−1 u −1 2x − 1 = + C = sec + C = sec +C
4
4
2
4
2
Z 3
t − 2t2 + 3t − 4
dt. [§7.6, p414, #78]
(5). Evaluate the integral:
t2 + 1
Solution:
2t − 2
t3 − 2t2 + 3t − 4
= (t − 2) + 2
∵
2
t +1
t +1
Z 3
Z
Z
2
t − 2t + 3t − 4
2t − 2
∴
dt = (t − 2) dt +
dt
2
t +1
t2 + 1
Z
Z
t2
t2
2t
2
= − 2t +
dt
−
dt
=
− 2t + ln(t2 + 1) − 2 tan−1 (t) + C
2
2
2
t +1
t +1
2
√
(tan−1 x)2
√
. [§7.6, p414, #97]
(6). Find the limit: lim+
x→0
x x+1
Solution:
√
√ 2
√
tan−1 ( x)
−1
√
(tan
x)
2 tan−1 ( x)
x(1+x)
( 00 )
( 00 )
√
lim
=
=
lim √
lim
√ √
x→0+
x→0+
x→0+ (3x + 2) x x + 1
x x+1
x + 1 + 2√xx+1
( 00 )
=
lim+
x→0
√ 1
x(1+x)
2
12x
√
√ +13x+2
2 x x+1
= lim+
x→0
2
2
√
=
=1
2
(12x2 + 13x + 2) x + 1
2