Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

192 0 CHAPTER 14 PARTIAL DERIVATIVES
After entering the (x, y) pairs into a calculator or CAS, the resulting least squares regression line through the points is
approximately y = 0.75136x + 0.01053, which we round toy= 0. 75x + 0.01.
(c) Comparing the regression line from part (b) to the equation y = ln b +ax with x = ln(L/ I b = e 0 ·0 1 ~ 1.01. Thus, the Cobb-Douglas production function is
p = bL"' ](1-a = 1.01£0.75 ](0.25.
14.2 Limits and Continuity
1. In general, we can't say anything about /(3, 1) I lim f(x, y) = 6 means that the values of f(x, y) approach 6 as
{x,y)-•(3,1) ·
(x, y) approaches, but is not equal to, (3, 1). Iff is continuous, we know that lim f(x, y) = f(a, b), so
(x,y)--z
-0.2
-0.1
-0.05
- 0.2 -0.1
-2.551 -2.525
-2.525 -2.513
-2.513 -2.506
- 0.05 0 0.05 0.1 0.2
-2.5 13 -2.500 -2.488 -2.475 -2.451
- 2.506 -2.500 -2.494 -2.488 - 2.475
-2.503 -2.500 -2.497 -2.494 -2.488
0
-2.500 -2.500
- 2.500 -2.500 -2.500 - 2.500
0.05
-2.488 - 2.494
- 2.497 -2.500 -2.503 - 2.506 - 2.5 13
0.1
-2.475 - 2.488
- 2.494 -2.500 -2.506 -2.513 - 2.525
0.2
- 2.451 -2A75
-2.488 - 2.500 - 2.513 - 2.525 - 2.551
As the table shows, the values off ( x, y) seem to approach - 2.5 as ( x, y) approaches the origin from a variety of different
directions. This suggests that lim f(x, y) = -2.5. Since f is a rational function, it is continuous on its domain. f is
(:t:,!J) - {0,0)
o b o o
de fi ne d at
bl' I th l' !( ) 0203 + 0302 - 5 5 'fy'
( 0, 0 ), so we can use d1rect su stltutwn to esta IS 1 at 1m x, y =
= --, ven mg
0
(:r.,.y)-->{0,0) 2 - 0 0 2
our guess.
5. f(x, y) = 5x 3 - x 2 y 2 is a polynomial, and hence continuous, so ' lim f(x, y) = f(l, 2) = 5{1) 3 - {1) 2 (2) 2 = 1.
(x,y)-{1,2)
7 . f( x, y ) . =
4 - xy . . 1 .., . d h . . d .
tS a ratwna ;unctiOn an encc contmuous on tts omam.
X 2 +3y 2
.
(2, 1) is in the domain off, so f is continuous there and lim f(x, y) = !(2, 1) = (~)~ ( 2 ~~~~ 2
{x,y) --< (2,1) +
2
7
9. f(x, y) = (x 4 - 4y 2 )/(x 2 + 2y 2 ). First approach (0, 0) along the x-axis. Then.f(x, 0) = x 4 /x 2 = x 2 for x =f. 0, so
f(x, y) --+ 0. Now approach (0, 0) along they-axis. For y =f. 0, f(O, y) = -4y 2 /2y 2 = -2, so f(x, y)--+ -2. Since f has
two different limits along two different lines, the limit does not exist.
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