高等数学竞赛极限与连续真题.docx

高等数学竞赛极限与连续真题

第一篇：高等数学竞赛极限与连续真题高等数学竞赛极限与连续真题x211x21.计算：lim2x22x0(cosxe)sinxx2x40(x4),析：1x1282x2111x2x40(x4)又cosxex[14123x0(x2)][1x20(x2)]x20(x2)22x211x2故lim2x22x0(cosxe)sinx10(x4)1444x0(x)21xx2188xlim2limx03212xsinx2x030(x2)sinx22x0(x)222xnlnnlnn)的值。2.计算求lim(nnlnnn（选自广东省大学生高等数学竞赛试题）nlnnlnn2lnn2lnnnlnn)=lim[(1)]析：lim（nnlnnnnlnnlnn1tt2t,则原式lim()e.令t0n1t1nnlnn2n111(1)n1)n23n111111111()析：S2n1232n32n1242n1111111112()=1232n242nn1n2nn3.计算：lim(11111)=(12nn111nnn最后一式是函数f(x)故limS2n11在[0，1]区间上的积分和（n等份，取右端点）1x1dxln201xn1又limS2n1lim(S2n)ln2nn2n11n11)因此lim(1(1)n23nn4.设lim2006，试求,的值。nn(n1)nnn1n析：=111n(n1)1(1)1(10())n0()nnnn,n11显然由条件知0；而lim,n1n0()n0,因此有10,且xx2n5.计算：lim(12)nn2n110,10,10,2006，故20051,20062006xxx(n)x(n)2xnxxn2)n121x析：(1)(1)(12xnn2n2n2xn2n24nnx易知:1ex，nx进行变量代换，令nxm,则当n时m,并且mx,对122nx2nnxxxxm2lim(1)(1)ex因此有lim1nmxmmn2nxx2nx)e.由夹逼原理得lim(12nn2n6.1.设当x1时,1m是x1的等价无穷小,则m______.1xxm1解m3.7.13.已知曲线yf(x)在点(1,0)处的切线在y轴上的截距为1,则lim[1f(1)]n_____.nn1解lim[1f(1)]ne.nn8.5.limnk1nkenn1k______.解原式e1.1.设函数yy(x)满足y(x1)yx2yex,且y(0)1.9.y(x)x若lima,则a_______.x0x2解应填1.由题设y(0)y(0)1,于是y(0)2.y(x)xy(x)11所以alimlimy(0)1.2x0x02x2x10.2.已知f(x)exb在xe处为无穷间断点,在x1处(xa)(xb)为可去间断点,则b________.解应填e.由题意知必有a1,be或ae,b1.e,limf(x),符合题意;x1e1xe当ae,b1时,limf(x)limf(x),与题意不符.x1xe当a1,be时,limf(x)11.7.已知lim1ln[1xx021f(x)f(x)]4,则lim_________.3x01cosxx解应填2ln2.1f(x)f(x)limln[1]limlimx0x02x11cosx(2x1)(1cosx)x0f(x)2f(x)f(x)lim4lim2ln2.2x0xln2x0x3x3xln2212.5.已知有整数n(n4)使极限lim[(xn7x42)x]存在且不为零,则__.x1.5因为lim[(xn7x42)x]lim[xn(17x4n2xn)x],所以由极限存在可得n1,解应填xx由极限不为零得4n1,因此