Thomas Calculus 13th [Solutions]

988 Chapter 14 Partial Derivatives
53. First consider the vertical line
along x 0
2 2
2x y 2(0) y
x 0 lim lim lim 0 0. Now consider any
4 2 4 2
( x, y) (0,0) x y y 0 (0) y y 0
nonvertical line through (0,0). The equation of any line through (0,0) is of the form y
Thus
along y mx
along y mx
2 2 3 3
2x y 2 x ( mx) lim f ( x, y) lim lim lim 2mx lim 2mx lim 2mx
0.
4 2 4 2 4 2 2
( )
2 2 2 2 2
( x, y) (0,0) ( x, y) (0,0) x y x 0 x mx x 0 x m x x 0 x x m x 0 x m
2x y
lim 0.
4 4
( x, y) (0,0) x y
anyline through (0,0)
54. If f is continuous at ( x0 , y 0),
then
2
lim f ( x, y ) must equal f ( x0 , y 0) 3. If f is not continuous at
( x, y) ( x , y )
( x0 , y 0),
the limit could have any value different from 3, and need not even exist.
0 0
mx
55.
2 2
lim 1 x y
( , ) (0,0)
3
1 and
x y
lim 1 1 lim tan xy
1
( x, y) (0,0) ( x, y) (0,0)
xy
1
by the Sandwich Theorem
56. If
x
2
y
2
x
2
y
2
xy
6 6
2 | xy| 2
xy 0, lim
( , ) (0,0)
| | lim lim 2 xy
xy
( , ) (0, 0)
xy
( , ) (0, 0)
6
2 and
2| xy|
lim
x y x y x y
( x, y) (0, 0)
| xy|
lim 2 2; if xy 0,
( x, y) (0,0)
x
2
y
2
x
2
y
2
xy
6 6
2| xy| 2
lim
( , ) (0, 0)
| | lim lim 2 xy
xy
( , ) (0,0)
xy
( , ) (0, 0)
6
2 and
x y x y x y
lim 2| xy|
4 4cos | |
( , ) (0,0)
| | 2 lim xy
xy
( , ) (0,0)
| xy|
2, by the Sandwich Theorem
x y
x y
57. The limit is 0 since sin 1 1 1 sin 1 1 y y sin 1 y for y 0, and y y sin 1
x x x
x
y for
y 0. Thus as ( x, y ) (0,0), both y and y approach 0 y sin 1 0, by the Sandwich Theorem.
x
58. The limit is 0 since cos 1 1 1 cos 1 1 x x cos 1 x and x x cos 1
y y y
y
x for x 0.
Thus as ( x, y ) (0,0), both x and x approach 0 x cos 1 0, by the Sandwich Theorem.
y
59. (a) f ( x, y ) 2m
2 tan
y mx
sin 2 . The value of f ( x , y ) sin 2 varies with , which is the lines
2 2
1 m 1 tan
angle of inclination.
(b) Since f ( x, y ) y mx sin 2 and since 1 sin 2 1 for every
1 along y mx.
, lim f ( x, y ) varies from
( x, y) (0,0)
1 to
60.
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
xy x y xy x y x y x y x y x y x y x y x y
2 2 2 2
2
2 2
2 2
2 xy x y x y
2 2 2 2
xy x y
2 2
x y x y x y x y
2 2 2 2 2 2
x y x y x y
2 2
y
2 2
x
lim xy 0 by the Sandwich Theorem, since
( x, y) (0,0) x y
f (0,0) 0
2 2
lim x y 0; thus, define
( x, y) (0,0)
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