Thomas Calculus 13th [Solutions]

326 Chapter 4 Applications of Derivatives
59.
2
y x 4 1 1
60.
2 2
x 3 x 3
y
2
x 1 4
2 2
x 4 x 4
61.
lim x 2 3x 4 2 3
1
1 lim x
x
1
1
5
x
x
62.
a
a 1
lim x 1 lim ax a
b
b 1
x 1 x 1 x 1 bx b
63. lim tan x tan 0
x
x
64.
lim tan x lim sec x 1 1
x 0
x sin x
x 0
1 cos x 1 1 2
2
65.
66.
67.
lim 2
sin x lim 2sin x cos x lim sin(2 x) lim 2cos(2 x)
2
2 2 2 2 2 2 2 2 2 2
0 tan( ) 0 2 sec ( ) 2 sec ( ) 0 2 (2sec ( ) tan( ) 2 ) 2sec ( ) 0 21
1
x x x x x x x x x x x x x x
sin( mx) m cos( mx)
lim lim
m
x 0
sin( nx) x 0
n cos( nx)
n
lim sec(7 ) cos(3 ) lim cos(3 x) 3sin(3 ) 3
cos(7 ) lim x
x x
x
7sin(7 x) 7
x /2 x /2 x /2
68. lim sec lim x
x x
0
cos x 1
0
x 0 x 0
69. lim (csc x cot x ) lim 1 cos x sin 0
0 0
sin lim x
x
0
cos x 1
0
x x x
70.
2
2 2
lim 1 1 lim 1 x lim (1 x ) 1 lim (1 x ) lim 1 1
4 2 4 4 4
x 0 x x x 0 x x 0 x x 0 x 0 x
71.
2 2
2 2 2 2 x x 1 x x
lim x x 1 x x lim x x 1 x x
lim 2x
1
2 2 2 2
x x x x 1 x x x x x 1 x x
Notice that
2
x x for x > 0 so this is equivalent to
2x
1
2
1
x
x
x
2
x 1 x
2
x x 1
1 1
1
1
2 2
x
x
2 x
x x
lim lim 2 1
x
1 1
72.
3 3 3 2 3 2
3 2
lim x x lim x ( x 1) x ( x 1) lim 2x lim 6x lim 12x
lim 12 1
2 2 2 2 4 3 2
1 1 ( 1)( 1) 1 4 12 24x
lim 2x
0
x x x x x x x x x x x x x x
73. The limit leads to the indeterminate form 0 0 : x
10 1 (ln10)10
lim lim ln10
x
x
x
1
x
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