1、 (1) sin 2limx xx→∞= 0 .(2) d(arctan )x = 21d 1+x x(3)21d sin x x =⎰ -cot +C x x(4).2()()x n e = 22n x e .(5)x =⎰26/32、(6) The right proposition in the following propositions is ___A_____.A. If lim ()x af x →exists and lim ()x ag x →does not exist then lim(()())x af xg x →+does notexist.B. If lim ()x af x →,lim ()x ag x →do both not exist then lim(()())x af xg x →+does not exist.C. If lim ()x af x →exists and lim ()x ag x →does not exist then lim ()()x af xg x →does notexist.D. If lim ()x af x →exists and lim ()x ag x →does not exist then ()lim()x af xg x →does not exist. (7) The right proposition in the following propositions is __B______.A. If lim ()()x af x f a →=then ()f a 'exists.B. If lim ()()x af x f a →≠ then ()f a 'does not exist.C. If ()f a 'does not exist then lim ()()x af x f a →≠.D. If ()f a 'does not exist then the cure ()y f x =does not have tangent at (,())a f a .(8) The right statement in the following statements is ___D_____.A. sin lim 1x xx→∞= B. 1lim(1)x x x e →∞+=C. 11d 1x x x C ααα+=++⎰ D. 5511d d 11b b a a x y x y =++⎰⎰ (9) For continuous function ()f x , the erroneous expression in the followingexpressions is ____D__. A.d (()d )()d b a f x x f b b =⎰ B. d (()d )()d baf x x f a a =-⎰C. d (()d )0d b a f x x x =⎰D. d (()d )()()d baf x x f b f a x =-⎰(10) The right proposition in the following propositions is __B______.A. If ()f x is discontinuous on [,]a b then ()f x is unbounded on [,]a b .B. If ()f x is unbounded on [,]a b then ()f x is discontinuous on [,]a b .C. If ()f x is bounded on [,]a b then ()f x is continuous on [,]a b .D. If ()f x has absolute extreme values on [,]a b then ()f x is continuous on [,]a b .3、Evaluate 2011lim()x x e x x →-- 201=lim()x x e x x →--01=lim()2x x e x →-01=lim =22x x e →（考点课本节洛比达法则，每年都会有一道求极限的解答题，大多数都是用洛比达法则去求解，所以大家要注意节的内容。

注意洛比达法则的适用范围。

）4．Find 0d |x y =and (0)y ''if 20x x x y y t e +=+⎰.2'()'x x x y y t e +=+⎰（）1'2()'2()1x x y x y x e y x y x e +=⋅+⇒=⋅+-0(20(0)1)0x dy y e dx dx==⋅⋅+-=''(2()1)'2()2'()x xy x y x e y x xy x e =⋅+-=++200-(0)0-01x x y y t e x y e =+⇒=+=⎰0''02(0)20'(0)=3y y y e =+⋅+（） （考察微积分基本定理与微分，书上节）5、 Find 22arctan d (1)xx x x +⎰=22221)arctan d (1)x x x x x x +-+⎰（22arctan arctan =d d (1)x x x x x x -+⎰⎰-12311=-arctan +d arctan +2x x x x x x -⎰22-1221++1=-arctan +d arctan 1+2x x x x x x x x -⎰（） -12211=-arctan +d d arctan 1+2x x x x x x x x --⎰⎰（）-12211=-arctan +In In 1+arctan 22x x x x x ---121=-arctan arctan +C 2x x x - （凑微分求不定积分，积分是微积分的重点及难点，大家一定要掌握透彻。

）6、 Given that 22()1x f x x =+.(1) Find the intervals on which ()f x is increasing or decreasing.22’22212()1x x x x f x x +-⋅=+（）（）2221x x =+（）When ’()00f x x >⇒>’()00f x x <⇒< Therefore, the increasing interval is ()0+∞，, the decreasing interval is ()0-∞，(2) Find the local maximum and minimum values of ()f x’()00f x x =⇒= The function is increasing in interval ()0+∞，, decreasing ininterval ()0-∞，, therefore, the function exist the local minimum value, it is ()0f x =(3)Find the intervals of concavity and the inflection points.'22224222242422181642''()111x x x x x x f x x x x +-+--+⎛⎫=== ⎪+++⎝⎭（）（）（）（）（）4224642''()0133x x f x x x x --+=>⇒><-+（）4224642''()0001x x f x x or x x --+=<⇒<<<<+（）1''(4f f =Therefore, the concave upward interval are 3⎛⎫-∞- ⎪ ⎪⎝⎭，,3⎛⎫+∞ ⎪ ⎪⎝⎭, the concave downward intervalare -03⎛⎫⎪⎪⎝⎭,03⎛⎫ ⎪ ⎪⎝⎭， and the inflection pointsare 134⎛⎫ ⎪ ⎪⎝⎭，,134⎛⎫⎪ ⎪⎝⎭， (4) Find the asymptote lines of the cure ()y f x =2221lim =1111+x x x x→∞=+ T herefore, the liney = 1 is a horizontal asymptote（考点：节，、节。

近几年经常会考一道作图题。

这种题目应该在注意的点主要包括函数的定义域，对称性，增减区间，极值点，凹凸性，拐点，以及渐近线等。

大家参照课本的节进行作图）7、Let R be the region bounded by the curve 1y x=, and the line y x = and 2x =.(a)Evaluate the area of the region R. R=211x dx x -⎰2211=In 2x x -2211=2In21In122⋅--⋅+3=In22- (b)Find the volume of the solid generated by revolving the R about they -axis .V=212121214)?4)y dy dy y ππ-+-⎰⎰（（12311211443x y y y π⎛⎫ ⎪=-++ ⎪⎝⎭33111142241141423312π⎛⎫=⋅-⋅-⋅+⋅+⋅+-⋅- ⎪⎝⎭83π=（考点：求面积以及体积，课本、节。

这类题目是常考题，较简单。

望同学一定要做相应的题目加以巩固。

）8、 Determine the production level that will maximize the profit for a company with cost and demand functions23()1450360.580.001C x x x x =+-+and ()600.01p x x =-.Solution 2()(600.01)600.01R x x x x x=-=-223()()()600.01(1450360.580.001)P x R x C x x x x x x =-=--+-+320.0010.57241450x x x =-++-'2()0.003 1.142P x x x =-++Let '()040020P x x or x =⇒==- since x>0, then x=400''()0.006 1.14P x x =-+When x=400 ''()0.006400 1.14 1.260P x =-⋅+=-<32(400)0.0014000.5740024400145035350P =-⋅+⋅+⋅-=Therefore, when the production level is 400 that will maximize the profit 35350 （考点：经济函数，课本节。