TOPIC IV CALCULUS. [1] Limit. Suppose that we have a function y = f(x). What would happen to y as x x? o

Transcription

1 [1] Limit TOPIC IV CALCULUS Suppose that we have a function = f(). What would happen to as? o Definition: Let = f(). Then, the limit value of as a is denoted b lim a f(). EX 1: = 1 +. As 0, 1. lim = 1. 0 EX : = 1/, 0. As 0, or - Limit does not eists. As 0 from the right,. 1 lim o. As 0 from the left, -. 1 lim 0. IV-1

2 1 EX 3: =. 1 lim = 1; lim 1 ;. 1 EX 4: =, lim1 = = 0 (?) 0 lim 1 lim = 0; lim- = 0. Nope!!!. 1 (1)(1) lim 1 = lim 1 = lim 1(1+) =!!! 1 1 EX 5: =, 0. lim0 = lim0 = 0. lim = lim =. Note: Before computing a limit, better to simplif the function f. Although f() ma not be defined at =, f() could have a limit. o IV-

3 a EX 6: =. a lim = lim (1 + ) = 1. Theorem: Suppose that lim f() and lim g() eists. (Here, lim = lim o.) Then, (R1) lim [f() ± g()] = lim f() ± lim g(). (R) lim[f()g()] = [lim f()][lim g()] (R3) lim[f()/g()] = [lim f()]/[lim g()], if lim g() 0. EX: = e /. lim = [lime ]/[lim ] (?) Nope!!!, since lim e and lim do not eist. In fact, lime / =. lim0 = [lim0e ]/[lim0 ] (?) Nope!!!, since lim0 = 0. In fact, lim =. 0 EX: = ( -1)/(-1), 1. lim1 = [lim ( -1)]/[lim (-1)] (?) Nope!!!, since lim (-1) = 0. 1 In fact, lim1 = lim 1(+1) =. IV-3

4 EX: = ( +)/( ++). lim = [lim ( +)]/[lim ( ++)] (?) Nope!!!, since lim lim = lim ( +) does not eists. 1/ 1// = [lim (1+/ )]/[lim (1+/+/ )] = 1. L'Hôpital's Theorem Suppose that we have two functions f() and g(). Suppose: lim f() = 0 and lim g() = 0, or, lim f() = ± and lim g() = ±. f() f () Then, lim = lim. g() g () IV-4

5 [] Derivative = f() We want to know changes in () when changes b. rate of change = f( 0 ) f( 0 ) 0 d d f f( ) f() () = lim. 0 d d = derivative evaluated at 0 = f( 0) 0 Note: If lim 0 lim, then, we sa that derivative does not eist. 0 Question: What is derivative? IV-5

6 =f() 0 / : tangent line at = 0 f( ) : measures slope of tangent line 0 IV-6

7 [3] Continuit vs. differentiabilit Definition: Consider a function = f(). Suppose: 1) f is defined at ; 0 ) limf() limf() limf() ; ) lim f() f( 0 ). 0 Then, f is called continuous at =. 0 EX 1) = f() = 6, for all. Is this function continuous at = 1? = 6 IV-7

8 1) f(1) = 6; ) lim 6; 1 3) lim 1 f(1) Continuous. EX ) 1 Is this function continuous at = 0 1) f(0) not defined. = 1/ is not continuous at = 0. IV-8

9 +1 =+ - 1 EX 3) = 1 if 1 if < 1 1) f(1) =. ) lim f() ; 1 Not continuous. lim f() 3 1 Definition: Suppose that at 0, f( ) = 0 lim 0 f( 0 ) f( 0 ) eists. IV-9

10 Note: Then, we sa that f is differentiable at. 0 If lim 0 lim, we sa that derivative does not eist. 0 Theorem: If f is differentiable at 0, then, f is continuous at 0. <proof> Note that: f() f( 0 ) f() - f( 0) = ( 0 ). 0 Then, lim 0 [f() f( 0 )] f() f( lim 0 ) = ( 0 0 ) = f( 0) 0 = 0. 0 Thus, lim f() f( 0 ) 0 lim f() = f( o). 0 0 Corollar: If f is not continuous, then f is not differentiable. EX 1) = 1 1 < 1 Not continuous at = 1 IV-10

11 not differentiable. Note: If a function is continuous, it ma or ma not be differentiable. Definition: a = a if a 0 a if a < 0 EX 1) -4 = 4; 4 = 4 EX) = f() = = 1 if 3 if < Continuous? ) lim f() lim f() 1 lim 0 1) f() defined and f() = 1 3) lim f() = f(). So, f is continuous. Differentiable? f( ) f() = ( ) 1 1 lim 0 = = 1 lim 0 IV-11

12 lim 0 f( ) f() = ( ) 3 1 lim 0 = = -1 lim 0 not differentiable IV-1

13 [4] Rules for Derivatives (1) = f() = k. d d 0. () = a n = an n-1 EX 1) = = 3 6 = 18 EX ) = = = d d (3) (f() g()) = d d f() d d g() f () g () EX) = + d d 4 1. IV-13

14 d (4) d (f()g()) = d d f() g() f() d d g() d d f() d f() g() f() d d g() (5) =. d g() [g()] EX) f() = g() = f() = 1 + 6; g() = d 3 f() g() = (1 + 6)(3 + 6 ) + (6 + 6)(9 + 1) d d f = d g (1 6)(3 3 6 ) (6 6)(9 1) (3 3 6 ) IV-14

15 [5] Chain Rule z z = f() = g() dz d dz d z w d dz d. dz d d d d f () g () EX 1) z = 3 ; = + 5 z dz d dz d = (6) = 1 = 1( + 5) = d d dz d EX ) = 3( +1). Set z = Then, = 3z ; z = + 1. d/d = (d/dz)(dz/d) = (9z )(4) = 36z = 36( +1). EX 3) TR = f(q); Q = g(l) dtr dl dtr dq dq = MR MP L= MRP dl IV-15 L

16 [6] Inverse Function (EX 1) Consider the following eample. This is a function. Looking at the reversed relation: (EX ) Consider the following eample. IV-16

17 Now, look at the reversed relation: Definition: If the reversed relation is also a function, we sa that there eists the inverse -1 function, and we denote it b f. Definition: A function = f() is strictl increasing (decreasing) iff f( 1) > f( ), whenever 1> (<). Theorem: -1 If a function = f() is strictl increasing or strictl decresing, then = f () (inverse function) eists. Note: IV-17

18 (1) Function : If ou know what happened esterda (cause), ou can predict what will happen toda (result). () Inverse function : If ou know what happens toda (result), ou can figure out what happened esterda (cause). EX) = f() = 5 + 5, - < <. --1 = /5-5 = f (). Theorem: Suppose that a function = f() has an inverse function. Then, d d 1 d d. EX) For the above eample, d/d = 5 ; d/d = 1/5. Clearl, d/d = 1/(d/d). IV-18

19 [7] Partial Differentiation Consider a function: = f( 1,,..., n), where i s are independent variables and is a dependent variables. Suppose we would like to know the rate of change of when 1 changes. Definition: d ceteris paribus 1 d 1 lim. 1 1 (Here, we treat all other variables as constant!!!!) EX 1) = / = / = EX ) = ( + 4)(3 + 4 ). 1 1 / = [( +4)/ ](3 +4 ) ( +4)[(3 +4 )/ ] = ( +4) 3 = / = (Show this at home.) 1 EX 3) 1 1. IV-19

20 1 ( 1 ) 1 ( 1 ) ( 1 ) ( 1 ) 1 ( 1 ) ( 1 ) 1 (1) ( 1 ) 3 ( 1 ). Simliarl, / = (1-1 )/(1- ). IV-0

21 [8] Jacobian Determinant There are n equations: 1 1 = f ( 1,,... n, n+1, n+,... m) = f ( 1,,..., n, n+1, n+,..., m) n n = f ( 1,,..., n, n+1, n+,..., m). Let s treat,... as given. (Treat them as constants) n+1 m Sometimes, we wish to know whether all the functions are distinctive, that is, whether there are redundant functions or not. For this case, we use Jacobian matri: J = n n n 1 n... n n If J = 0, we sa that the functions are functionall dependent. If J 0, we sa that the functions are functionall independent. IV-1

22 EX 1) ; ; ; 1 1 J EX ) = = ( 1 + ) J ( 1 4 ) 0 IV-

23 [9] Differentials d : derivatives d d, d : differentials (small changes) d d f () d f ()d EX) = d d f () 6 7 d = (6 + 7)d One important application of differentials: (price-elasticit of demand) d = % changes in demand % changes in price = % changes in demand when price changes b 1% Q d Q d dq d Q d = P dp = P P dq dp P Q d EX) Q = P d IV-3

24 dq d dp P Q d P 100 P = d () P 100 P P P When P = 10, d = = Total differentiation: = f(,... ) 1 n Denote f j j Then, d = f d + f d f d 1 1 n n d f 1 d 1. d d n 0 EX) U = U(, ) 1 du U d 1 U d 1 = U d + U d 1 1 IV-4

25 Suppose that du = 0. Then, U d + U d = d d U 1 (slope of indifference curve) du 1 0 U EX) d = ( + )d + ( + 8 )d IV-5

26 [10] Total derivatives CASE I: = f(, ) ; = g(w) ; = h(w) 1 1 d 1 d 1 d EX 1) 1 3 ; 1 = 6w + 6; = 3w + 4w d 1 d 1 d = ( 1 3 )6 (3 1 )(6w 4) = 1(6w+6)(3w +4w) + 3(6w+6) (3w +4w) (6w+4) = 7(w+1)(3w +4w) + 108(w+1) (3w +4w) (6w+4). EX ) 3 = (6w+6) (3w +4w) Set 1 = 6w + 6; = 3w + 4w. d 1 d 1 d IV-6

27 CASE II: = f(,, w) 1 = g(w) 1 = h(w) d f 1 d 1 f d f w Note: d/ = total derivative. f/w = partial derivative from EX 1) 1 w ; 1 = 6w; = 3w d 1 d 1 d w = ( ) w + 1 = 1(6w) + 6w + 1 = 78w IV-7

28 EX ) = 3 - w ; = w + w + 4 d Show that = 10w + 3. Total Partial Derivatives: = f(,, z, z) 1 1 = g(z, z) 1 1 = h(z, z) 1 d dz 1 Want to know (in tetbook, ) dz 0 z 1 z 1 f 1 f f z 1 1 z 1 z 1 z 1 fr. fr. fr. fr. fr. EX 3) w = a + b + cu; = u + v; = u IV-8

29 w u f u f u f u. = (a + b) + (b) + c = a + b + b + c w v w v = (a + b) = w u du w v dv IV-9