Part 10
題庫
 x 7 3
, x2

P4-11. Suppose that f ( x)   x  2  2
is continuous at x  2 , then k  ?

k,
x2

(sol): For f ( x) to be continuous at x  2 , we must have lim f ( x)  f (2) . Thus
x 2
lim f ( x)  f (2)
x 2
 lim
x 2

x  2  2 
 x  7   9  
 lim
 x  2   4  
x 7 3
 k  lim
x 2
x2 2


x7 3
x2
 lim
x2
 x  2  2   k
x  2  2  x  7  3
x  2  2
k
x  7  3
x7 3
x22
42
4
2
k
k kk 
6
3
x7 3
9 3
＃
P4-19. lim n 3n  4n =?
n
(sol): the indeterminate form 0 and lim r n  0 if 1  r  1
n 
Since,
4n  3n  4n  2  4n  n 4n  n 3n  4n  n 2  4n
1
 4  n 3n  4 n  2 n  4
Further, lim 4  4 and
n
lim 1n
1
lim 2 n  4  4  2n  4  20  4 1  4
n 
So, from the Sandwich Theorem, we have that
l i mn n3
n
n
4
4
＃
1
P4-20. lim
x 


x 2  1  x 2  4 x =?
(sol): the indeterminate form   


 lim
x 
x2  1 
 

2
x2  1 

x
 lim
x2  4x

2
x2  1  x2  4 x
x 
x 
x2  1  x2  4x
2
x2  1  x2  4 x
x 
 lim

x  4 x 
x2  1  x2  4x
lim
2
 1   x 2  4 x 
x2  1  x2  4 x
 lim
4x 1
x2  1  x2  4x
4x
4x
 lim
 lim
x 
x 2  x 2 x  x  x
4x
4x
 lim
 lim
 lim 2  2
x  x  x
x  2 x
x 
＃
x 
 x3  1
x 1
 x 1 ,

1  x  2 , then
P5-1. Suppose that f ( x)  2 x  1,
 x3  3x  2

, 2 x
 x  2
(sol): For f ( x) to be continuous at x  1 , we must have lim f ( x)  lim f ( x) . We compute
x 1
x 1
the one-sided limits. Thus
 x  1  x 2  x  1
x3  1
lim f ( x)  lim
 lim
 lim  x 2  x  1  12  1  1  3
x 1
x 1 x  1
x 1
x 1
x 1
and
lim f ( x)  lim  2 x  1  2 1  1  3
x1
x1
Since the one-sided limits equal the value of the function at the point x  1 , f ( x) is
continuous at x  1 . Similarly, we verify that the function f ( x) is also continuous at the
point x  2 or not. However,
lim f ( x)  lim (2 x  1)  2  2  1  5
x  2
x 2
and
2
 x  2   x 2  2 x  1
x3  3x  2
lim f ( x)  lim
 lim
x  2
x 2
x 2
x2
x2
2
 lim  x  2 x  1  22  2  2  1  9
x 2
from which we get that lim f ( x ) does not exist. Hence, f is discontinuous at x  2 .
x2
＃
 kx 2  x  4k  2
, x2

P5-2. Let f ( x)  
be continuous at x  2 , then k  ?
x2

3,
x2

(sol): For f ( x) to be continuous at x  2 , we must have lim f ( x)  f (2) . Thus
x 2
lim f ( x)  f (2)
x 2
 x  2  kx  2k  1  3
kx 2  x  4k  2
 3  lim
x 2
x 2
x2
x2
 lim
 lim  kx  2k  1  3  2k  2k  1  3  4k  2  k 
x 2
P5-3. The function f is defined by f (h) 
1
2
＃
(2  h)2  4
, determine which one is wrong in
h
the following statements.
(sol): Here, f is a rational function and is continuous for h  0 . Thus, f is continuous at
h  1 . The value of f (0) is undefined, thus, f (0) does not exist. In addition, the limit
(2  h) 2  4
4  4h  h 2  4
4h  h 2
lim f (h)  lim
 lim
 lim
 lim  4  h   4  0  4
h 0
h 0
h 0
h 0
h 0
h
h
h
Since, f (0) does not exist. Hence, f is discontinuous at h  0 .
＃
 x2  4
, x2

P6-8. Suppose that f ( x)   x  2
is a continuous function, then k  ?
 k, x  2

(sol): For f ( x) to be continuous at x  2 , we must have lim f ( x)  f (2) . Thus
x 2
3
lim f ( x)  f (2)
x 2
 lim
x 2
 x  2  x  2   k
x2  4
 k  lim
x 2
x2
x2
 lim  x  2   k  2  2  k  k  4
x 2
＃
3k  x , 0  x  4
P6-12. Suppose that f ( x)  
is continuous on [0, 9] , then k  ?
2kx  7 x, 4  x  9
(sol): For f ( x) to be continuous at x  4 , we must have lim f ( x)  lim f ( x) . Thus
x 4
x 4
lim f ( x)  lim f ( x)
x  4
x 4


 lim 3k  x  lim  2kx  7 x   3k  4  2k  4  7  4  5k  30  k  6
x 4
x 4
＃
cos x, x  0

P6-14. Suppose that f ( x)  a  x 2 , 0  x  1 is a continuous function, then a  b =?
bx,
1 x

(sol): For f ( x) to be continuous at x  0 , we must have lim f ( x)  lim f ( x) . We
x 0
x 0
compute the one-sided limits. Thus


lim f ( x)  lim cos x  cos 0  1 and lim f ( x)  lim a  x 2  a  0  a
x  0
x 0
x 0
x 0
 a 1
Similarly, the function f ( x) is also continuous at the point x  1 . Hence,
l i mf x( )
x 1
l i am (2x  )
x 1
x 1
2
l
im
x ( 1 2 ) 1and1 lim2f ( x)  lim bx  b
x 1
x 1
b  2
Therefore,
a  b  1  2 3
＃
P7-18. If the interval I includes at least a real root for the equation x3  15 x  1  0 , find
the interval I .
(sol): f ( x)  x3  15x  1 , do the following table
4
x
4
3
2
1
0
1
2
3
4
f ( x)
3
19
23
15
1
13
21
17
5
Thus
f (4)  f (3)  3 19  57  0
f (0)  f (1)  1 (13)  13  0
f (3)  f (4)  17  5  85  0
The function f changes sign on intervals [4, 3] , [0,1] and [3, 4] . Hence there must be
a zero on each interval.
＃
x2  2 x  3
=?
x  2 x 2  5 x  1
(sol): the indeterminate form  
P8-5. lim
x2  2 x  3
x2
1 1

lim
 lim 
2
2
x  2 x  5 x  1
x  2 x
x  2
2
＃
lim
1  2 x3
=?
x  x  1
(sol): the indeterminate form  
P8-6. lim
1  2 x3
2 x3
 lim
 lim  2 x 2   
x  x  1
x 
x 
x
＃
lim
n!
=?
n  n n
(sol): n !  1 2  3 
P8-9. lim
 (n  1)  n
Since,
0 
n! 1 2 3
   
nn
n n n
n

1n

n
n

1
n
1
 0 . So, from the Sandwich Theorem, we have that
n  n
Further, lim 0  0 and lim
n
n!
limn  0
n  n
＃
x  cos x
=?
x 
x
P9-13. lim
5
(sol): the indeterminate form   , we have that
1  cos x  1
We multiply above inequality through by 1 x ( x  0 ), we get

1 cos x 1


x
x
x
Further,
1
 1
lim     0  lim
x 
x

x
 x
So, from the Sandwich Theorem, we have that
c o sx
lim
 0
x 
x
This implies that
x  c o sx
c oxs
 x c o sx

lim
 l im 
  x l im 1
   1 0
x 
x


x
x 
x 
 x

＃
P9-15. The rational function f ( x) 
1
x2  x
horizontal asymptote is
x2  2x  3
(sol): Horizontal Asymptote
x2  x
l i mf x( ) l i 2m

1
x 
x  x  2 x  3
 y  1 is a horizontal asymptote of the graph of f .
＃
P9-16. The rational function f ( x) 
x2  x
vertical asymptote is
x2  2x  3
(sol): Vertical Asymptote
lim
f x( )

x 3
x2  x
l i 2m

x 3 x  2 x  3

x 3
(x x 1 )
lim
 
x(  3x)(
1)
and
x2  x
(x x 1 )
l i m

lim
 
2

x 1
x 1 x  2 x  3

x
1 x
(  3x)(
1)
 x  3 and x  1 are vertical asymptotes of the graph of f .
lim
f x( )

＃
kx  1, x  3
P9-17. f ( x)  
is continuous on
2  kx, x  3
, then k  ?
(sol): For f ( x) to be continuous at x  3 , we must have lim f ( x)  lim f ( x) . Thus
x 3
6
x 3
lim f ( x)  lim f ( x)
x 3
x 3
 lim  kx  1  lim  2  kx   3k  1  2  3k  6k  1  k 
x 3
x 3
7
1
6
＃