高等數(shù)學(xué)微積分課件多元函數(shù)極限的連續(xù)性

1、Chapter 8: Functions of Several VariablesSection 8.2Limits and ContinuityWritten by Richard GillAssociate Professor of MathematicsTidewater Community College, Norfolk Campus, Norfolk, VAWith Assistance from a VCCS LearningWare GrantThis section will extend the properties of limits and continuity fro

2、m the familiar function of one variable to the new territory of functions of two or three variables. I hate to bring up painful memories but here is the formal definition of a limit back when we were dealing with functions of one variable.)( then ,0if that so 0 a exists there0each for that means )(l

3、imstatement The number. real a be Llet and c)at possibly (except c containing intervalopen an on definedfunction a be fLet LxfcxLxfcxIn less formal language this means that, if the limit holds, then f(x) gets closer and closer to L as x gets closer and closer to c.ccc( )LLL( )x is the inputf(x) is t

4、he outputJust to refresh your memory, consider the following limits. ?044)2(2242lim222xxxGood job if you saw this as “l(fā)imit does not exist” indicating a vertical asymptote at x = -2.?004)2(2242lim222xxxThis limit is indeterminate. With some algebraic manipulation, the zero factors could cancel and r

5、eveal a real number as a limit. In this case, factoring leads to4121lim)2)(2(2lim42lim2222xxxxxxxxxThe limit exists as x approaches 2 even though the function does not exist. In the first case, zero in the denominator led to a vertical asymptote; in the second case the zeros cancelled out and the li

6、mit reveals a hole in the graph at (2, ).xy42)(2xxxfThe concept of limits in two dimensions can now be extended to functions of two variables. The function below uses all points on the xy-plane as its domain.3),(22yxyxfzxyzIf the point (2,0) is the input, then 7 is the output generating the point (2

7、,0,7).(2,0)(2,0,7)If the point (-1,3) is the input, then 13 is the output generating (-1,3,13).(-1,3)(-1,3,13)For the limit of this function to exist at (-1,3), values of z must get closer to 13 as points (x,y) on the xy-plane get closer and closer to (-1,3). Observe the values in the table to see i

8、f it looks like the limit will hold.13),(lim?)3 , 1(),(yxfyxThe concept of limits in two dimensions can now be extended to functions of two variables. The function below uses all points on the xy-plane as its domain.3),(22yxyxfzxyz(2,0)(2,0,7)(-1,3)(-1,3,13)For the limit of this function to exist at

9、 (-1,3), values of z must get closer to 13 as points (x,y) on the xy-plane get closer and closer to (-1,3). Observe the values in the table to see if it looks like the limit will hold.(x,y)(x,y,z)(0,0)(0,0,3)(-1,1)(-1,1,5)(-1,2)(-1,2,8)(-1,2.5)(-1,2.5,10.25)(-1, 2.9)(-1, 2.9, 12.41)(-.9,3)(-.9,3,12.

10、81)(-1.1,3)(-1.1,3,13.21)13),(lim?)3 , 1(),(yxfyxThe table presents evidence that the limit will hold, but not proof. For proof we have to go back to epsilon and delta.3),(22yxyxfzxyzDefinition of a Limit.L-y)f(x, then )()(0ifsuch that 0 a exists there0every for that means ),(limstatement The itself

11、. b)(a,at possibly except b),(a,center with circle a ofinterior thethroughout defined be variables twoof ffunction aLet 22),(),(byaxLyxfbayx13),(lim?)3 , 1(),(yxfyxIn the context of the limit we examined,suppose that .25.If the limit holds, we should be able to construct a circle centered at (-1,3)

12、with as the radius and any point inside this circle will generate a z value that is closer to 13 than .25.Center (-1,3)(x,y)25.13),(yxfDefinition of Continuity of a Function of Two VariablesA function of two variables is continuous at a point (a,b) in an open region R if f(a,b) is equal to the limit

13、 of f(x,y) as (x,y) approaches (a,b). In limit notation:).,(),(lim),(),(bafyxfbayxThe function f is continuous in the open region R if f is continuous at every point in R.The following results are presented without proof. As was the case in functions of one variable, continuity is “user friendly”. I

14、n other words, if k is a real number and f and g are continuous functions at (a,b) then the functions below are also continuous at (a,b):0b)g(a, if ),(),(),(/ ),(),(),(),(),(),( ),(),(yxgyxfyxgfyxgyxfyxfgyxgyxfyxgfyxfkyxkfThe conclusions in the previous slide indicate that arithmetic combinations of

15、 continuous functions are also continuousthat polynomial and rational functions are continuous on their domains.Finally, the following theorem asserts that the composition of continuous functions are also continuous.).,(),(lim and b)(a,at continuous is),(),)(function n compositio then theb),f(a,at c

16、ontinuous is g and b)(a,at continuous is f If),(),(bafgyxfgyxfgyxfgbayxExample 1. Find the limit and discuss the continuity of the function.yxxyx2lim)2, 1(),(Solution21412) 1 (212lim)2, 1(),(yxxyxThe function will be continuous when 2x+y 0.Example 2. Find the limit and discuss the continuity of the

17、function.yxxyx2lim)2, 1(),(Solution412) 1 (212lim)2, 1(),(yxxyxThe function will be continuous whenThe function will not be defined when y = -2x. . 02 yxxyzExample 3. Use your calculator to fill in the values of the table below. The first table approaches (0,0) along the line y=x. The second table a

18、pproaches (0,0) along the line x=0. (If different paths generate different limits, the official limit does not exist.) Use the patterns to determine the limit and discuss the continuity of the function.)ln(2122yxz)ln(21lim22)0, 0(),(yxyx(x,y) z(2,2)(1,1)(.5,.5)(.1,.1)(.01,.01)(x,y) z(0,2)(0,1)(0,.5)(0,.1)(0,.01)0.35-1.04-0.351.964.26-0.6900.692.304.61Calculate these values yourself. Then click to confirm.xyz)ln(2122yxz)ln(21lim22)0, 0(),(yxyx(x,y) z(2,2)(1,1)(.5,.5)(.1,.1)(.01,.01)(x,y) z(0,2)(0,1)(0,.5)(0,.1