1、 (1) sin 2limx x x→∞= 0 . (2) d(arctan )x = 21d 1+xx(3)21d sinx x=⎰-cot +Cx x(4).2()()x n e = 22n x e .(5)0x =⎰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 not exist.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 not exist.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 lim1x x x→∞= B. 1lim (1)x x x e →∞+=C.11d 1x x xCααα+=++⎰D.5511d d 11b b aax yxy=++⎰⎰(9) For continuous function ()f x , the erroneous expression in the following expressionsis ____D__. A.d (()d )()d ba f x x fb b =⎰ B.d (()d )()d baf x x f a a =-⎰C.d (()d )0d baf 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 211lim ()xx e xx→--21=lim ()xx e xx→--01=lim ()2xx e x→-01=lim=22xx e→（考点课本4.4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4.4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 2x xx y y t e+=+⎰.2'()'x xx y y t e +=+⎰（）1'2()'2()1xxy x y x e y x y x e +=⋅+⇒=⋅+-(20(0)1)0x dyy e dx dx==⋅⋅+-=''(2()1)'2()2'()xxy x y x e y x xy x e=⋅+-=++2-(0)0-01x x y y t e x y e =+⇒=+=⎰''02(0)20'(0)=3y y y e =+⋅+（）（考察微积分基本定理与微分，书上5.3节）5、 Find 22arctan d (1)x x x x +⎰=22221)arctan d (1)x x xxx x +-+⎰（22arctan arctan =d d (1)x x x xxx -+⎰⎰-12311=-arctan +d arctan +2x x x x x x-⎰22-1221++1=-arctan +d arctan 1+2x xx 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 xx ---121=-arctan +Inarctan +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()1xf x x =+.(1) Find the intervals on which ()f x is increasing or decreasing. 22’22212()1x x x xf x x +-⋅=+（）（）2221xx =+（）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''()000133x x f x x or x x --+=<⇒-<<<<+（）1''(334f f =-Therefore, the concave upward intervalare 3⎛-∞-⎪⎝⎭，,3⎛⎫+∞ ⎪ ⎪⎝⎭, the concave downwardinterval are -03⎛⎫ ⎪ ⎪⎝⎭,03⎛⎫ ⎪ ⎪⎝⎭，and the inflection points are 1-34⎛⎫ ⎪ ⎪⎝⎭，,134⎛⎫⎪ ⎪⎝⎭，(4) Find the asymptote lines of the cure ()y f x =2221lim=1111+x xx x→∞=+T herefore, the liney = 1 is a horizontal asymptote（考点：4.3节，4.5、4.6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4.5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 the y -axis .V=21212121 4) y dy dyyππ-+-⎰⎰（（12311211443x yy yπ⎛⎫ ⎪=-++⎪⎝⎭33111142241141423312π⎛⎫=⋅-⋅-⋅+⋅+⋅+-⋅- ⎪⎝⎭ 83π=（考点：求面积以及体积，课本6.1、6.2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 withcost 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4.8节。