Thomas Calculus 13th [Solutions]

Chapter 2 Practice Exercises 107
38. From the graph we see that lim k( x) lim k( x)
x 0 x 0
so k cannot be extended to a continuous function at
a 0.
39. (a) f ( 1) 1and f (2) 5 f has a root between 1 and 2 by the Intermediate Value Theorem.
(b), (c) root is 1.32471795724
40. (a) f ( 2) 2 and f (0) 2 f has a root between 2 and 0 by the Intermediate Value Theorem.
(b), (c) root is 1.76929235424
41.
lim 2x
3 lim 2
x 2 0 2
x
x
x 5
x
5 7 7 5 0 5
3
42.
3
2
2
2
lim 2x
3 lim x 2 0 2
x 5x
7
5 0 5
2 7
x 5
x
2
43.
44.
2
lim x 4x
8 lim 1 4 8
3 3
0 0 0 0
2 3
x 3x x
x 3x 3x
1
1
x
2
2 7 1
7x
1 x 1
x
x
2
lim lim 0
1 0 0
0
x x
45.
lim
x
2
x 7 x lim x 7
x 1
x 1
1
x
46.
4 3
lim x x lim x 1
x 12x
128
3 128
x 12
x
3
47. lim sin x lim 1 0 since x as x lim sin x 0.
x x x x x x
48. lim cos 1 lim 2 0 lim cos 1 0.
49.
sin x 2
x x
sin x
1
x
lim
1
x sin x 2 x
1 0 0
x sin x
lim 1 0
1
x
x
50.
51.
2/3 1 5/3
lim x x lim 1 x 1 0
1 0
1
x x
2/3 2
cos
2
cos x x
x
1
x
2/3
1/ x 1 0
lim e cos e cos(0) 1 1 1
x
x
52. lim ln 1 1 ln1 0
x
t
53.
lim tan
x
1
x
2
Copyright
2014 Pearson Education, Inc.