微积分英文版4汇编

1、三、其他未定式三、其他未定式 二、二、 型未定式型未定式一、一、 型未定式型未定式00机动 目录 上页 下页 返回 结束 洛必达法则 第三三章 一、一、0)(lim)(lim) 1xFxfaxax)()(lim)3xFxfax存在 (或为 )()(lim)()(limxFxfxFxfaxax,)()()()2内可导在与axFxf0)( xF且定理定理 1.型未定式型未定式00(洛必达法则) 机动 目录 上页 下页 返回 结束 ( 在 x , a 之间)证证: 无妨假设, 0)()(aFaf在指出的邻域内任取,ax 则)(, )(xFxf在以 x, a 为端点的区间上满足柯0)(lim)(lim

2、) 1xFxfaxax故)()()()()()(aFxFafxfxFxf)()(Ff)()(limxFxfax)()(limFfax)()(limxFxfax)3定理条件定理条件: 西定理条件,机动 目录 上页 下页 返回 结束 )()(lim)3xFxfax存在 (或为 ),)()()()2内可导在与axFxf0)( xF且推论推论1. 定理 1 中ax 换为, ax,ax,xx之一,推论推论 2. 若)()(limxFxf满足定且型仍属)(, )(,00 xFxf理1条件, 则)()(lim)()(limxFxfxFxf)()(limxFxf 条件 2) 作相应的修改 , 定理 1 仍然成

3、立.,x)()(lim)()(limxFxfxFxfaxax洛必达法则定理1 目录 上页 下页 返回 结束 例例1. 求.123lim2331xxxxxx解解: 原式 lim1x型00266lim1xxx23注意注意: 不是未定式不能用洛必达法则 !266lim1xxx166lim1x332x1232 xx机动 目录 上页 下页 返回 结束 例例2. 求.arctanlim12xxx解解: 原式 limx型00221limxxx1211x21x11lim21xx型机动 目录 上页 下页 返回 结束 二、二、型未定式型未定式)(lim)(lim) 1xFxfaxax)()(lim)3xFxfax

4、存在 (或为)()(limxFxfax定理定理 2.证证: )()(limxFxfax仅就极限存在的情形加以证明 .)()(limxFxfax(洛必达法则)机动 目录 上页 下页 返回 结束 ,)()()()2内可导在与axFxf0)( xF且1)0)()(limxFxfax的情形)()(limxFxfax limax)(1xF)(1xf limax)()(12xFxF)()(12xfxf)()()()(lim2xfxFxFxfax)()(lim)()(lim2xfxFxFxfaxax)()(lim)()(lim1xfxFxFxfaxax)()(lim)()(limxFxfxFxfaxax从而

5、型00机动 目录 上页 下页 返回 结束 2)0)()(limxFxfax的情形. 取常数,0k,0 kkxFxfax)()(lim)()()(limxFxFkxfax)()()(limxFxFkxfax)()()(limxFxFkxfaxkxFxfax)()(lim)()(lim)()(limxFxfxFxfaxax可用 1) 中结论机动 目录 上页 下页 返回 结束 3)()(limxFxfax时, 结论仍然成立. ( 证明略 )说明说明: 定理中ax 换为之一, 条件 2) 作相应的修改 , 定理仍然成立., ax, ax,xx,x定理2 目录 上页 下页 返回 结束 例例3. 求. )

6、0(lnlimnxxnx解解:型原式11limnxxxnnxxn1lim0机动 目录 上页 下页 返回 结束 例例3. 求. )0(lnlimnxxnx解解:型原式11limnxxxnnxxn1lim0机动 目录 上页 下页 返回 结束 . )0(0lnlimnxxnx例3. 例4. )0, 0(0limnexxnx说明说明:例如,xxx21lim21limxxxxxx21lim而xxx21lim11lim2xx1用洛必达法则在满足定理条件的某些情况下洛必达法则不能解决 计算问题 . 机动 目录 上页 下页 返回 结束 3.1 Maxima & Minima Maxima: point

7、 whose function value is greater than or equal to function value of any other point in the interval Minima: point whose function value is less than or equal to function value of any other point in the interval Extrema: Either a maxima or a minimaWhere do extrema occur? Peaks or valleys (either on a

8、smooth curve, or at a cusp or corner)f(c)=0 or f(c) is undefined Discontinuties Endpoints of an interval These are known as the critical points of the function Once you know you have a critical point, you can test a point on either side to determine if its a max or min (or maybe neitherjust a leveli

9、ng off point)3.2 Monotonicity and ConcavityLet f be defined on an interval I (open, closed, or neither). Then f isINCREASING on I if, DECREASING on I if, MONTONIC on I if it is ether increasing or decreasing)()()()(21212121xfxfxxxfxfxxMonotonicity TheoremLet f be continuous on an interval I and diff

10、erentiable at every interior point of I.If f(x)0 for all x interior to I, then f is increasing on Ia) If f(x)0 for all x in (1,c) and f(x)0 for all x in (c,b), then f(c) is a local max. value.If f(x)0 for all x in (c,b), then f(c) is a local min. value.a) If f(x) has the same sign on both sides of c

11、, then f(c) is not a local extreme value.Second Derivative Test Let f and f exist at every point in an open interval (a,b) containing c, and suppose that f(c)=0.a)If f(c)0, then f is a local min. value of f.3.4 Practical Problems Optimization problems finding the “best” or “least” of “most cost effe

12、ctive”, etc. often involves finding the extrema of the function Use either 1st or 2nd derivative testExample A fence is to be constructed using three lengths of fence (the 4th side of the enclosure will be the side of the barn). I have 120 yd. of fencing and the barn is 150 long. In order to enclose

13、 the largest possible area, what dimensions of fence should be used? (continued on next slide)Example continued Area is to be optimized: A = l x w Perimeter = 120 yd = 360=2l + w w = 360 2l So, 2 lengths of 90 and a width of 180. HOWEVER, the barn is only 150 wide, so in order to enclose the greates

14、t area, we wont use a critical point of the function, rather we will evaluate the area using the endpoints of the interval, with w=150. The length = 55 and the area enclosed = 8250 sq. ft.180)90(236090,4360043602360)2360(2wlllAreallllArea3.5 Graphing Functions Using CalculusCritical points & Inf

15、lections points If f(x) = 0, function levels off at that point (check on either side or use 2nd deriv. test to see if max. or min.) If f(x) is undefined: cusp, corner, discontinuity, or vertical asymptote (look at behavior and limits of function on either side) If f(x) = 0: inflection point, curvatu

16、re changes3.6 Mean Value Theorem for Derivatives If f is continuous on a closed interval a,b and differentiable on the open interval (a,b), then there is at least one number c in (a,b) where )( )()()( )()(abcfafbfcfabafbfExample: Find a point within the interval (2,5) where the instantaneous velocit

17、y is the same as the average velocity between t=2 and t=5.sec5 . 3144)( )(sec1425)5)2(2(5)5(225)2()5(52)(222tttstvftssvttsaveIf functions have the same derivatives, they differ by a constant. If F(x) = G(x) for all x in (a,b), then there is a constant C such that F(x) = G(x) + C for all x in (a,b).3

18、.7 Solving Equations Numerically Bisection Method Newtons Method Fixed-Point AlgorithmBisection Method Let f(x) be a continuous function, and let a and b be numbers satisfying ab and f(a) x f(b) 0. Let E denote the desired bound for the error (difference between the actual root and the average of a

19、and b). Repeat steps until the solution is within the desired bound for error. Continue next slide.Bisection MethodnnnnnnnnnnnnnnnnnnnnbbandmasetmfafIfmbandaasetmfafIfabhCalculateSTOPmfifmfCalculatebamCalculate1111, 0)()() 5, 0)()() 42/ )(:) 3., 0)(),(:) 22/ )(:) 1Newtons Method Let f(x) be a differ

20、entiable function and let x(1) be an initial approximation to the root r of f(x) = 0. Let E denote a bound for the error. Repeat the following step for n = 1,2, until the difference between successive error terms is within the error.)( )(1nnnnxfxfxxFixed-Point Algorithm Let g(x) be a continuous func

21、tion and let x(1) be an initial approximation to the root ro of x = g(x). Let E denote a bound for the error (difference between r and the approximation). Repeat the following step for n 1,2, until the difference between succesive approximations are within the error.)(1nnxgx3.8 Antiderivatives Definition: We call F an antiderivative of f on an interval if F(x) = f(x) for all x in the interval.Power RuleCrxdxxrr11Integrate = Antidif