发布时间 : 星期日 文章南开大学数学竞赛练习题08sxjs1更新完毕开始阅读79c93237a32d7375a41780ef
(8)设函数f(x) =limlnex?xnnn????, 则f(x)的定义域是 . (1996北京竞赛本科)
n??111(9)xn?1?1?1?1?2???1?2???n,则limxn= .(1998北京竞赛本科)
(10)若数列{xn}满足条件lim(xn?1?xn)?0,则limn??n??xnn= .
(11)当x?0时,?(x)=kx2与?(x)?1?xarcsinx?cosx是等价无穷小,则k= . 1(12)lim(cosx)x?0ln(1?x2)= .(2003研招一)
(13)limx?0ln(1?sin2x)3?1xf(x)?5,则limfx(2x)= .(1996北京竞赛本科)
x?01n?1(14)设数列{xn}满足:nsin(15)lim(n??n?xn?(n?2)sin1n?1,则limn??1n?1?xk?1nk= .(2000北京竞赛本科)
a?nbn2)= ,其中a?0,b?0为常数,且a?1,b?1. (2002北京竞赛本科甲、乙)
f(x)x2(16)已知函数f(x)在x=0的某邻域内有定义,且limx?0?1,则limf(1?x?0f(x)x)= .
1x(17)若limatanx?b(1?cosx)2x?0ln(1?2x)?c(1?e?x)2=2,则a= .(2000北京竞赛本科甲、乙)
(18)曲线y=2xx?1的斜渐近线方程为 . (2005研招一) (19)lim(x?y)x?0,y?022xy= .
2)= .(2004北京竞赛本科甲、乙) (20)设曲线y= f(x)与y=sinx在原点相切,则极限limnf(nn??1.2.单项选择题:
?x2,x?0?2?x,x?0(1)设g(x)??, f(x)??,则g[f(x)]? ( )
?x?2,x?0??x,x?0?2?x2,x?0?2?x2,x?0?2?x2,x?0?2?x2,x?0. (B)?. (C)?. (D)?. (A)?2?x,x?02?x,x?02?x,x?02?x,x?0????(2)若f(x)与g(x)互为反函数,则f[g (3x)?2]的反函数是 ( ) (A) g[f (3x)?2] . (B) f[2g(x)]? 3 . (C) g[2 f(x? 3)] . (D)2g[f(x)? 3] . (3)定义在(??,??)上的下列函数中没有反函数的是 ( )
(A)y=x?sinx . (B) y=x?sinx . (C) y=xsinx . (D)y?(1?x)sgnx . (4)若对一切实数x ,都有f(x)= ?f(x?5) ,则曲线y= f(x) ( )
(A)向左(或向右)平移10个单位后与原曲线重合 . (B)关于直线x=5? 2对称 . (C)关于点(5? 2,0)对称 . (D)关于直线y= x对称 . (5)下列函数中的非周期函数是 ( )
(A)cos(sin?x?) . (B)(sinx). (C)sin(sinx) . (D)e2sinx2?e?sinx.
(6)设数列{xn}与{yn}满足limxnyn?0 ,则下列断言正确的是 ( )
n??(A)若{xn}发散,则{yn}必发散. (B)若{xn}无界,则{yn}必有界.
(C)若{xn}有界,则{yn}必为无穷小. (D)若{x1}为无穷小,则{yn}必为无穷小.(1998研招二)
n(7)若lim?(x)?u0且limf(u)?A,则 ( )
x?x0u?u0(A)limf[?(x)]存在 . (B)limf[?(x)]?A . (C)limf[?(x)]不存在 .
x?x0x?x0x?x0(D)A、B、C均不正确 . (2001天津竞赛理工)
1(8)曲线y=ex2?x?1arctan(xx?1)(x?2)的渐近线有 ( )
2(A)1条 . (B)2条 . (C)3条 . (D)4条 . (2002天津竞赛理工) (9)曲线y=x?x2?x?1 ( )
(A)没有渐近线 . (B)有一条水平渐近线和一条斜渐近线 .
(C)有一条铅直渐近线 . (D)有两条水平渐近线 . (2006天津竞赛理工) (10)设函数f(x)在(?∞,+∞)内单调有界,{xn}为数列,下列命题正确的是 ( ) (A)若{xn}收敛,则{f(xn)}收敛. (B)若{xn}单调,则{f(xn)}收敛.
(C)若{f(xn)}收敛,则{xn}收敛. (D)若{f(xn)}单调,则{xn}收敛 . (08研招一) 1.3.设常数a?0,求证:若f(x?a)=1?f?x?,或f(x?a) = 1.4.求函数f(x)=x?[x] , x????,??), 的反函数f?11?f?x?1?f?x?1?f?x?，则f(x)是周期函数.
?x? .
1.5. 解下列函数方程: 当x?0,1时, f(x)满足方程f(x)? f(1?1? x)=1?x .
1.6.给定下列函数f(x),定义f0(x)?f(x),fn?1(x)?f[fn(x)],n?N.求fn(x):
1 (1)f(x)?1?x?1?x . (2)f(x)?2xx??1.1.7.设对每一对实数(x, y) ,函数f满足方程f(x)?f(y)=f(x?y)?xy?1, f(1)=1 .求整数n使f(n) = n≠1 .
1.8.设定义在??? ,???上的函数f(x)满足f(x?T) = k f(x)，x???? ,???，其中T与k?1均为给定的正的常数,证明: 有正数a与以T为周期的函数?(x)使f(x) = a?(x), x???? ,??? .
1.9.试构造整系数多项式ax?bx?c,使它在(0,1)内有相异实根,且a是满足条件的最小正整数 . 1.10.证明: 若f(x)是单调增加函数，其反函数f1(n!)n. 1.11.计算极限: limnn??1x2?1(x) ? f(x)，则f(x) ? x .
n1.12.设limxn?a?0且a?0,求limxn.
x?an??n??{1.13.设limxn?a ,试讨论n??xn?1xn}的收敛性 .
n1.14.设xn??k?1n1k1 ?lnn .(1)求证数列{xn}收敛;(2)求lim?k?n .n??k?11.15.证明下述各数列{xn}收敛,并求limxn:
n??(1)x0?a,xn?1?1k?1?kxn?bxnk? ,n?N,给定正整数k及正的常数a与b .
n1(2)x0?3,xn?1?10?3xn,n?N . (3)x0?1 2,xn?1?1?x,n?N .(4)0?x1?3,xn?1?(5)x0?,xn?1?x0?a2xn(3?xn)(n?1,2,?).
xn22,n?N ,0?a?1.
,n?1,2,?,对每一个实数a讨论数列{xn}的收敛性,并在数列
1.16.设x1?0,xn?1?a?xn1?xn收敛时求出limxn.
n??1.17.设x0?1,x1?2,求下述各数列的极限：
(1)xn?2?12(xn?xn?1) , (2)xn?2?xn?1xn , (3)xn?2?n??2xnxn?1xn?xn?1 , n?N.
1.18.设0?x1?1,xn?1?xn(1?xn),n?1,2,?.求limxn与lim(nxn).
n??1.19.求证:(1)limn???k?0n1k!1?e; (2)e??k!?k?0n?nn?n!,其中0??n?1.
1.20.设a1?1,a2?2,当n?3时,an?an?1?an?2,证明:
1(1)32an?1?an?2an?1;(2)liman=0 .
n??1.21.设0?a0?a1???an??,xn?1.22.计算下列各极限:
?(1-k?1nak?1ak)1ak,n?1,2,? . 证明数列{xn}收敛 .
x(1)lim[ln(1?2)ln(1?)]. (2)lim[lnxln(1?x)]. (3)lim(?2). (4)lim(1?sin?n).
xx???3xx?1?x??1x1xnn??(?1)csc(?(5)lim[1?n]n??n1?n2).
(6)limx?0[sinx?sin(sinx)]sinxx4. (08研招一)
1.23.设f(x)=4x?3x,求:
2x?x（1）limf(x);(2)limf(x);(3)limf(x);(4)limf(x);(5)limf(x).
x???x???x?0?x?0?x?01.24.如果f(x)是(??,??)上的周期函数,且limf(1x)=0 . 求f(x) .
x?01.25.设x→0+时函数f(x)是与x等价的无穷小量,且f(x)>x,试求极限lim3x?0?[f(x)]x?xxf(x)?x.
1.26.设a? b是三维空间R上的两非零常向量?且|b|=1?∠(a? b)=π∕3?求极限
lim[1x (|a+xb|–|a|)].
x?0x?1.27.设f(x)在x=0的某邻域内有连续导数?且lim(sinx2x?0f(x)x)?2,试求f(0)及f ’(0) .
（三）习题解答或提示
21.1. (1){x|1<|x|<2}.(2)1? 2 . (3)(k? , k? ), k?Z .(4) k.(5)y?sgnxarcsinx,x?[?1,1] .
(6)1+x. (7)(b?a)? 4 . (8)(- 1,1). (9)5 ? 2 . (10) 0 . (11)3 ? 4 . (12)e.(13)10ln3 .
121(14)1. (15)ab. (16) e . (17) ?4 . (18)y?1 (19)1 . (20) 2x?4.2?122 .
1.2. D ,B,C,A,C, D ,D,B,B.
?arctanx?1,x?0?0,x?0 ,xn?(10)B正确. A的反例是f(x)=??arctanx?1,x?0?(?1)nn. C与D的反例是f(x)=arctan x ,
xn=n . 应选: B .
1.3.证.(1)由f(x?a) =1?f(x)解得f(x)?1?f?x?a??f(x?2a),故f(x)以2a为周期 . (2)由f(x?a)=1?f(x)得f(x)??1.4.f?11?f(x)1?f?x?a?1?f(x)1f(x?2a)f(x) = f(x?4a), 故f(x)以4a为周期 . ,f(x?2a)??f(x1?4a),?x?=x?m , 2m?x?2m?1, m?Z . 1.5.
12nf(x)?x3?x2?12x(x?1).
??2,x??21n?n?11.6.(1)fn(x)??2x,?21n?x??2,1x?2n? , n∈N .(2)f6(x)?f0(x),f6n?k(x)?fk(x),0?k?5.
1.7.解.令y = 1，从原方程可得f(x?1) = f(x) ? x ? 2 ?????????????????? ?1?
由此递推式看出x为正整数时，f(x)为正，且f(x?1) ? x ? 2? x?1，再从0与负整数中求解，将 x = 0, ?1, ?2等代入?1?的等价式 f(x) = f(x?1) ?(x?2) ?????????????????? ?2?
得f(0) = ?1，f(?1) = ?2，f(?2) = ?2，于是有一解n = ?2 . 此外还有f(?3) = ?1，f(?4) = 1 .当x??4时，?(x?2)?2，由?2?式可知f(x) ? 0，不可能有f(x) = x .总之我们得唯一整数解n= ?2 . 1.8.证. 由k?0，T ?0，可设a=k1T?0,其中a?1,我们有f(x?T)?aTf(x) .现令