Key to Problem Set 2
Instructor: Dr. Ming Xu
Due Date: Sept. 27, 2004
P127:4~29
4
lim ( x 2  3 x  7)  (2) 2  3(2)  7  9
x  2
5 lim (t 3  5t 2  4)  (0 3  5(0) 2  4  4
t 0
6 lim ( x  5)( 2 x  7)  (3  5)( 2 * 3  7)  8
x 3
e  2 x  sin 1 x e  2*(0 )  sin 1 (0) 1  n


 1  n , n is integer.
x 0
x 1
0 1
1
z 2  z  3 12  1  3
lim

 1 / 2
x 1
z 1
11
x 2  3 x  10
3 2  3 * 3  10
lim

 8 / 35
x 3 3 x 2  5 x  7
3 * 32  5 * 3  7
4  u2
(2  u)( 2  u)
lim
 lim
 lim (2  u)  2  (2)  4
x  2 2  u
x  2
x  2
2u
x2  4x  4
( x  2)( x  2)
x2 22
lim 2
 lim
 lim

0
x 2 x  x  2
x  2 ( x  2)( x  1)
x 2 x  1
2 1
1
1
 ( x  1)
1 1
lim x
 lim
 lim

 1
x 1 x  1
x 1 x ( x  1)
x 1 x
1
7 lim
8
9
10
11
12
( x  1) 2  1
( x  1  1)( x  1  1)
x ( x  2)
 lim
 lim
 lim ( x  2)  0  2  2
x 0
x 0
x 0
x 0
x
x
x
x 1
x 1
1
1
14 lim
 lim
 lim

 1/ 2
x 1 x  1
x 1
( x  1)( x  1) x 1 x  1 1  1
13 lim
15 lim
y 2
y2 2
y2 2
y2 2
 lim
 lim
2
y 2 y  2  2
y 2
y2
( y  2  2)( y  2  2)
 lim
y 2
1
y22

1
22 2
 1/ 4
1
sin 2 x
sin 2 x
 lim (
* 2)  1 * 2  2
x 0
x 0
x
2x
sin 4 x
sin 4 x 4
4
17 lim
 lim (
* )  1*  4 / 9
x 0
x

0
9x
4x
9
9
cosh  1
18 lim
0
h 0
h
1  cos x
cos x  1  x
x
19 lim
 lim (
*
)  0 * lim
0
x 0
x

0
x

0
sin x
x
sin x
sin x
20 lim e  x sin x  e 0 sin 0  0
16 lim
x 0
21 lim (ln x) cos x  (ln 1) cos 1  0
x 1
2,000
0
x   x  1
7,000
lim
0
x  
x 1
3x  5
lim
3
x   x  2
x2
lim
 1/ 3
x   3 x  5
x
lim
1
x  
x 2  1,000
22 lim
23
24
25
26
3x
 3/ 2
4 x 2  10
x 5.916  1 5.916
28 lim

1
x  
35
x 35
x 6.083  1 6.083
29 lim

1
x  
37
x 37
27 lim
x  
2
P127:34~39
34 lim
x
x 0
x
Solution :
x
 1
x 0 x
x 0
x
x
x
lim
 lim  1
x 0 x
x 0 x
x
x
x
So, lim
 lim . That is , lim
doesn' t exist.
x 0 x
x 0 x
x 0 x
lim
x
35 lim
 lim
x2
x2
Solution :
x  2
x2
 ( x  2)
 1
x  2 x  2
x  2
x2
x2
x2
lim 
 lim 
1
x  2 x  2
x  2 x  2
x2
x2
x2
So, lim 
 lim 
. That is , lim
doesn' t exist.
x  2 x  2
x  2 x  2
x  2 x  2
lim 
 lim 
36 Solution:
1,000t
 1,000.
t 
t  20  t
So, if t approaches infinity, y(t) has its limiting value.
lim y(t )  lim
37 Solution:
3,000
3,000e 0.35t

lim
 3000,
t  
t   1  300e  0.35t
t   e 0.35t  300
3,000
lim N (t )  lim
0
t  
t   1  300e  0.35t
So, lim N (t ) doesn' t exist. But for t  0, we get lim N (t )  3000
(1) lim N (t )  lim
t 
t  
1
3,000
lim N (t )  1500 
 1500  1  300e 0.35t  2
 0.35t
t


2
1  300e
ln 300
 300e 0.35t  1  t 
 16
0.35
( 2) N ( t ) 
3
38 Solution :
20t
 0 for t  0
e 0.5 t
lim C (t )  lim 20te 0.5 t  lim
t 
t 
t 
Graph:
From the graph, we find when t=2, C (t ) is maximal.
39 Example :
f ( x) 
x
x
g( x)  
, lim f ( x )  lim
x 0
x
x
x 0
x
x
x
, lim g ( x )  lim 
x 0
x 0
But lim [ f ( x )  g ( x )]  lim [
x 0
doesn' t exist
x 0
x
x
x
doesn' t exist
 (
x
x
)]  lim 0  0
x 0
P137; 4~6,12~14,18~19,32~34
4. the number of unemployed people in the United States during January 1997 considered
as a function of time.
Solution: this is not a continuous function because once a person is employed, there will
be a jump of the curve. However, if the number of people employed equals the number of
people laid off all the time, it could be a continuous function.
Domain: 0  t  31, t is integer .
5. the charges to mail in a package as a function of its weight.
Solution: it isn’t a continuous function because this function could be
A, x  (0, c]
f (x) 
{ B , x  (c, d ]
C , x  (d ,)
4
6. the number of bacteria in colony as a function of time.
Solution: it is not a continuous function because any death or birth could give rise to a
jump of the curve. However, if the number of death equals the number of birth all the
time, it could be a continuous function since the number of bacteria will be constant.
Domain: t  0
12 Solution:
Suspicious points: t  1, t  3
lim g (t )  lim (3t  2)  g (1)  3 * 1  2  5
t 1
t 1
lim g (t )  lim 5  g (1)  5
t 1
t 1
 lim g (t )  g (1)  5
t 1
lim g (t )  lim 5  g (3)  5
t 3
t 3
lim g (t )  lim (3t 2  1)  3 * 3 2  1  26  g (3)
t 3
t 3
 lim g (t ) doesn' t exist, and the function at t  3 is not continuous .
t 3
13 Solution:
g ( x )  e x and h( x )  x are both continuous functions. So, f(x) 
g(x) e x

are continous
h(x)
x
when x  0. The suspicious point is x  0
ex
ex
lim f(x)  lim
 , lim f(x)  lim
 
x 0 x 0 - x
x 0 
x 0  x
 lim f(x)  lim f(x)
x 0 -
x 0 
 f ( x ) breaks at x  0.
14. Solution:
g ( x )  ln x and h( x )  x  1 are both continuous functions. So, f(x) 
g(x) ln x

are continous
h(x) x  1
when x  0 and x  1. The suspicious point is x  1
ln x
ln( u  1)
ln x
ln( u  1)
 lim
 1, lim f(x)  lim
 lim
1
u

0
x

1

x

1

u

0

x 1
u
x 1
u
 lim f(x)  lim f(x)  lim f(x )  1
lim f(x)  lim
x 1-
x 1-
x 1-
x 1
x 1
 f ( x ) is a hole at x  0.
5
18 Solution:
lim f ( x )  lim
x 2
cos
x 2

x  lim
x  2 x 2
sin(



)
2 x  lim
x 2
x2
sin
 ( x  2)
2x
x2
 lim [
x 2
sin
 ( x  2)
2x
 ( x  2)
*

2x
] /4
2x
 we can define that f ( x )   / 4 when x  2
19 Solution:
lim f ( x )  lim (15  x 2 )  15  4  9
x 2
x 2
lim f ( x )  lim (2 x  5)  2 * 2  5  9
x 2
x 2
 lim f ( x )  9
x 2
 we can define that f ( x )  9 when x  2
32 Solution:
let p( x )  ln x  ( x  2) 2
p(1)  ln 1  (1  2) 2  1
p(2)  ln 2  (2  2) 2  ln 2
 p(1)  0  p(2)
 we can find at least one solution w hich makes p( x )  0 when x  [1,2]
33 Solution:
lim f ( x)  lim ( x  1)  3  f (2)
x 2 
x 2 
lim f ( x)  lim x 2  4  f (2)
x 2 
x 2 
 f is continuous from the left at 2, but not from the right.
34 Solution
f (2)  2a  b, f (0)  0  b  b, f (2)  f (0)
 2a  b  b
a0
lim f ( x )  lim ( x 2  b)  1  b
x 1
x 1
lim f ( x )  lim (ax  b)  a  b  b
x 1
x 1
lim f ( x )  lim f ( x )  f ( x )  3
x 1
x 1
 1 b  b  3
 No such constants meet the standard.
6
P138: 37~38, 40, 44~47
37 Solution:
(a) lim E(v)  lim
v v w
v v w
C (v w  v ) k
(v  v ) k  (v w ) k  (v w ) k
Cv k
 lim
 C * lim w
v 0
v  v w v 0
v
v
(v w  v ) k  (v w ) k
(v ) k
 C lim w , v  v  v w
v 0
v  0 v
v
d
 C ( [ x k ]) | x v w  Ck (v w ) k 1  
dx
 Ck (v w ) k 1    
Interpretation: when a fish swims upstream with a speed approaching the speed of water,
the speed of the fish approaches 0 relative to the ground. Under this circumstance, the
fish can’t get to a point upstream no matter how much energy it expenses.
Cv k
(a) lim E(v)  lim
 
v 
v  v  v
w
Interpretation: when a fish swims upstream with an extremely high speed relative to the
water, it needs infinite energy to travel to a point upstream. That is to say, higher speed
the fish swims, more energy it needs.
 C lim
38 Solution:
(a) The colony die out  P(t )  0
if t  5  -8t  66  0  t  8.25 minutes
if 0  t  5  t 2  1  0  t doesn' t exist
 t  8.25 minutes
(b) P(t  2)  2 2  1  5 thousands  5,000
P(t  7)  8 * 7  66  10 thousands  10,000
 P(2)  9,000  P(7)
 the population is 9,000 at some time between t  2 and t  7
40 Solution:
(a) 0.638 * 365  232.87 days
Since there are 212 days from January to July, the birthday of this individual must be in
August 20th, 1981, which is derived from 232-212=20. And his/her birthday must be in
the time period of 16:00~24:00 of that day, which comes from 0.87*24= 20.88.
7
(b) Discontinuous function:
y (t )  t  a
Where
y(t) is the year when the individual was born;
t is the current year and is an integer;
a is the current age of the individual .
(c) Continuous function:
y (t )  a  kt
Where
y (t ) is the height of the individual ;
t is the time after the individual was born, t  R;
a is the innitial height of the individual when she/he was born;
k is a constant.
44. Solution:
lim f ( x )  lim
x 0 
x 0 
x3  4x
( x  2)( x  2)
 lim
 lim ( x  2)  2  f (0)  b
2
x 0 
x2
x  2 x x 0 
b2
lim f ( x )  lim (e x  a )  1  a  f (0)  2
x 0 
x 0 
 a  1
45 Solution
sin ax
 a  f ( 0)  5
x 0 
x
lim f ( x )  lim
x 0 
a2
lim f ( x )  lim ( x  b)  b  f (0)  5
x 0 
x 0 
b5
46 Solution:
lim f ( x)  lim 2 sin( a cos 1 x)  2 sin( a (n1 
x 0 
x 0 
 a (n1 
a

2
)  2n 2 
2n 2 
n1 

3 or


3
or 2n 2 
2n 2 
n1 
2
3

2
))  f (0)  3
2
3 , n and n are both integer .
1
2

2
2
lim f ( x)  lim (ax  b)  b  f (0)  3
x 0 
x 0 
b 3
8
47 Solution:
If f (a ) * f (c )  0  the root lies in [a,c]
If f (b) * f (c )  0  the root lies in [c,b]
9