Exam 3 Exam 3                              Friday, May 11, 2001



NAME:

Student Number:------            No.----           Marks ----






Instructions:

1. You have 50 minutes for this exam. You are not allowed to use any calculators.

2. This exam contains 2 questions and has a total of 100 marks.

3. Show all of your work. Your work must justify the answer that you give.

1 (40 Marks) Find the following limit



lim
x 0+ 

sinx
x

1/x2
 
.

Solution. Let f(x) = ([sinx/x])[1/(x2)]. Then we have lnf(x) = [(ln([sinx/x]))/(x2)] = [(lnsinx-lnx)/(x2)].



lim
x 0+ 
lnf(x)
=

lim
x0+ 
lnsinx-lnx
x2
=
lim
x 0+ 
(lnsinx-lnx)
2x
=

lim
x 0+ 
xcosx-sinx
2x2sinx
=
lim
x 0+ 
cosx-xsinx -cosx
2(2xsinx+x2cosx)
=

lim
x 0+ 
-sinx
2(2sinx+xcosx)
=
lim
x 0+ 
-cosx
2(3cosx-xsinx)
=
-1/6.

Hence limx 0+lnf(x) = -1/6. Since f(x) = elnf(x) and g(x) = lnx is continuous, we have



lim
x 0+ 
f(x) =
lim
x 0+ 
elnf(x) = elimx0+lnf(x) = e-1/6.


2.     (60 Marks)     Evaluate each of the following integrals.

(a)    x[(5x+3)] dx

(b)    e2xsin(2x) dx

(c)    [1/(x3-5x2+8x-4)] dx

Solution. (a) Let u = 5x+3. Then x = (u-3)/5 and dx = du/5. Hence, we have



x   ____
5x+3
 
 dx
=
1
25

(u-3)u du
=
1
25

(u3/2-3u1/2 du
=
1
25

2
5
u5/2-2u3/2
+C = 2
125
u5/2- 2
25
u3/2+C
=
2
125
(5x+3)5/2- 2
25
(5x+3)3/2+C

(b)



e2xsin(2x) dx
=
1
2


sin(2x) de2x
= 1
2

e2xsin(2x)-
e2xcos(2x) d(2x)
=
1
2

e2xsin(2x)-
cos(2x) d(e2x)
=
1
2

e2xsin(2x)-e2xcos(2x)+
e2x d cos(2x)
=
1
2

e2xsin(2x)-e2xcos(2x)-2
e2xsin(2x) dx
.

This implies 2e2xsin(2x) dx = [1/2][e2xsin(2x)-e2xcos(2x)]+C and



e2xsin(2x) dx = 1
4

e2xsin(2x)-e2xcos(2x)
+C.

(c) Since [1/(x3-5x2+8x-4)] = [1/((x-1)(x-2)2)], we let


1
(x-1)(x-2)2
= A
x-1
+ B
x-2
+ C
(x-2)2
.

Multiplied by (x-1)(x-2)2, we obtain


A(x-2)2+B(x-1)(x-2)+C(x-1) = 1.

Let x = 1. Then A = 1.

Let x = 2. Then C = 1.

Let x = 0. Then 4A+2B-C = 1. This implies B = -1. Hence, we have


1
(x-1)(x-2)2
= 1
x-1
- 1
x-2
+ 1
(x-2)2
.

Integrating both sides of the above equation, we obtain



1
(x-1)(x-2)2
 dx
=

1
x-1
 dx-
1
x-2
 dx+
1
(x-2)2
 dx
=
ln|x-1|-ln|x-2|- 1
x-2
+C.



File translated from TEX by TTH, version 2.79.
On 12 May 2001, 09:32.