Thomas Calculus 13th [Solutions]

118 Chapter 3 Derivatives
32.
(2 ) (2)
lim f h f
h 0
h
lim
h
0
4 3 4 3
(2 h) (2)
3 3
h
lim
h
0
4
2 3
3 [12 h 6 h h ]
h
4
0
3
lim [12 6 h h ] 16
h
2
( ( ) ) ( ) mh
h 0
( x0 h)
x0
0
h
0
0 0
33. At ( x0 , mx0
b ) the slope of the tangent line is
The equation of the tangent line is 0
lim m x h b mx b
h
y ( mx b)
m( x x0
) y mx b.
lim lim m m .
h
34. At x 4, y
1 1
and m
4
2
lim
h
0
1 1
4 h 2
h
lim
h 0
1 1
h 2
4 2 4
h 2 4
h
h
lim
h 0
2 4
2h
4
h
h
lim
h 0
2 4 h 2 4 h
2h 4 h 2 4 h
4 (4 h)
lim
lim
h
h 0 2h 4 h 2 4 h h 0 2h 4 h 2 4 h
lim
h 0
1
2 4 h 2 4
h
1 1
2 4 2 4 16
(0 ) (0)
35. Slope at origin lim f h f
h 0
h
origin with slope 0.
2
sin
1
h
h
lim lim h sin 1 0 yes, f ( x ) does have a tangent at the
h 0
h
h 0
h
36.
(0 ) (0)
lim g h g
h 0
h
1
h
h sin
lim lim sin 1 . Since lim sin 1
h 0
h
h 0
h
h 0 h
does not exist, f ( x ) has no tangent at the origin.
37.
(0 ) (0)
lim f h f
1 0
(0 ) (0)
lim , and lim f h f
h
h
h
h 0
h 0
h 0
yes, the graph of f has a vertical tangent at the origin.
1 0
(0 ) (0)
lim . Therefore, lim f h f
h
h 0
h 0
h
38.
(0 ) (0)
lim U h U
0 1
(0 ) (0)
lim , and lim U h U
h
h
h
h 0
h 0
h 0
vertical tangent at (0, 1) because the limit does not exist.
lim 1 1 0
h
h 0
no, the graph of f does not have a
39. (a) The graph appears to have a cusp at x 0.
(b)
(0 ) (0)
lim f h f
2/5
lim h 0 lim 1 and 1
h
h
3/5
3/5
h 0
h 0 h 0 h
hlim
0 h
2/5
of y x does not have a vertical tangent at x 0.
limit does not exist
the graph
40. (a) The graph appears to have a cusp at x 0.
(b)
(0 ) (0)
lim f h f lim h 4/5 0
h
h
h 0
h 0
hlim
0 h
does not have a vertical tangent at x 0.
1
1/5
and 1
1/5
hlim
0 h
limit does not exist
y
4/5
x
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