=========================preview======================
(math023)[2006](f)final~PPSpider^_10401.pdf
Back to MATH023 Login to download
======================================================
Math 023 -Calculus I
Final Examination, Fall Semester, 2006

Time Allowed: 3 Hours Total Marks: 155
Note: Calculators are not allowed. Please write details to receive marks.

1. Calculate the following limits: tan x . x
(a) (5 Marks) lim ;
x0 x(1 . cos 2x)
Solution : tan x . x sec2 x . 1
lim = lim
x0 x(1 . cos 2x) x0 1 . cos 2x +2x sin 2x
2 sec2 x tan x
= lim
x0 4 sin 2x +4x cos 2x tan x
2 sec2 x = lim x
x0 sin 2x
4+4 cos2x
x
2
= =1/6.
8+4
xx .
2
(b) (5 Marks) limn arctan . arctan ;
n1 nn +1
Solution If x = 0, then the limit is 0. If x 6
= 0, we replace t = n, with t being a continuous variable, and apply LHospitals rule: xx .
lim n 2 arctan . arctan
n1 nn +1 arctan t.1x . arctan(t + 1).1x = lim
t1 t.2 (.1)t.2x . (.1)(t+1).2x 1+t.2x2 1+(t+1).2x2
= lim
t1 (.2)t.3
12t4x + t3x
= lim = x.
t1 2(t2 + x2)(t2 +2t + x2 + 1)
Thus for any real x,
xx .

lim n 2 arctan . arctan = x.
n1 nn +1
sin(xx . 1)
(c) (10 Marks) lim .
x0+ x ln x . x
Solution : First,

xx ln x
lim x = lim e.
x0+ x0+ By LHospitals rule, it is easy to get limx0+ x ln x = 0. Thus, lim x x = 1. This gives
x0+
sin(xx . 1) sin(xx . 1) xx . 1 xx . 1
lim = lim = lim .
x0+ x ln x . x x0+ xx . 1 x ln x . x x0+ x ln x . x For the later limit, we use LHospitals rule and have xx . 1 xx(1 + ln x)
lim =lim =1.
x0+ x ln x . x x0+ ln x
Hence,

sin(xx . 1)
lim =1.
x0+ x ln x . x
.x2/2
cos x . e
(d) (10 Marks) lim .
x0 x4
Solution We have the taylor expansions:
24
xx
cos x =1 . ++ o(x 4),
2! 4!
2 ! 2 !2 2 !3
1 x
.x2/2 xx
e = 1+ . + . + o .
2 2! 2 2
24
xx
=1 . ++ o(x 6).
28
Thus,

.x2/2
cos x . e
lim 4
x0 x
24 24
1 . x+ x+ o(x4) . 1+ x. x+ o(x6)
2! 4! 28
= lim = .1/12.
x0 x4
x d3y
2. (a) (5 Marks) For y = ln .... ...., .nd ;
x2 . 1 dx3
Solution Since y = ln |x|. ln |x . 1|. ln |x +1|, we have
0
y = x .1 . (x . 1).1 . (x + 1).1 . Hence y 000 =2x .3 . 2(x . 1).3 . 2(x + 1).3 . d2y
t
(b) (10 Marks) For x = cos e,y =1 . sin et, .nd ;
dx2
Solution
dy ttt
dy . cos e ecos e
dt
== = .
dx
dx (.1) sin et et sin et
dt
Hence,
d dy

d2y dx
=
dx2 dx
t
d cos e sin et
,dx
=
dt dt
t tt
. sin2 e et . cos2 e e1
=
sin2 et (.1) sin et et 1
= .
sin3 et
x
(c) (10 Marks) For the curve y = y(x) de.ned by xy + y= 2, .nd the tangent line equation at (1, 1).
Solution By the implicit di.erentiation, we have
yx.1 x
yxy.1 + x ln x y 0 + xy y 0 + y ln y =0.
0
At (1, 1), we have y= .1. Hence, the tangent line is y . 1= .(x . 1),
i.e., y = .x + 2.
3. (15 Marks) Given .1
. x 3 sin ,x = 0; 6
f(x)= x. 0,x =0.
Show that f0(x) is continuous at x = 0, but f00(0) does not exist.
Solution If x 6
= 0, then 11
f0(x)=3x 2 sin . x cos .
xx
At x = 0,

f(h) . f(0) h3 sin 1 . 01
h
f0(0)= lim =lim =lim h2 sin =0.
h0 h h0 h