大学高数第一章习题课

结本章小数（会求定义域，函数值）一,函limfx求（（）=A）法二,极限1)分析定义2)夹逼性3)单调有界必有极限4)四则运算()()0()fxAfxAx6)（反映了极限与无穷小的关系，常用于理论证明）7)注意：乘除可用，等价加量代换（减不能）两个重要8)极限公式00(0),(0)fxfx分段点----必要9)须求1limsin0xxx无穷小性质（无穷小量乘以有界量仍为无穷小，例5)）(1)1sinlim0xxx(2)exxx)11(limexxx10)1(lim00limxxfxfx（）=（连（）续）三,1,初等函数的连续区间----定义域----即讨论讨论分段函数的端点,分连续性段点连续性000000000000limlimlimlimlimlimxxxxxxxxxxxxfxfxfxfxfxfxfx可去（）（）第一类跳跃（）（）（）第二类（无穷---）或（）或2,间断点（分类）3,闭区间上连续函数性质(4个定理)2(1)log(16.)xyx求函数的定义域解,0162x,01x,11x214xxx,4221xx及(1,2)).(2,4即X0-4412例)cos1,)xxxf(sin求f(cos22()?afaf(x)为奇函数，且x0当时，f(x)=lnx,则当a0时222cos21sinxx（）22)xf(sin2222()21,(cos)21cos22sin22xxfttxf（）（）f(x)为奇函数f(-x)=-f(x)()[()]afaafaa0时,ln()aa例例解解2ff(x)=1-x+ln(1-2sinx),求D21012sin0xx111sin2xx16x1261-1XY例解例.设,0)(,1)]([,)(2xxxfexfx且求)(x及其定义域.由得,)1ln()(xx]0,(xxx)(x解:)(x例已知8,)]5([8,3)(xxffxxxf,求.)5(f][f解:)5(f)(f)10(f)7(f][f)(f)9(f6例设,coscsc)sin1(sin22xxxxf求.)(xf解:1sin)(sin2sin1sin12xxfxx3)(sin2sin1xx3)(2xxf111()1xfxxf(x)=,求1111xyxxy=y11()1xfxx1111()1111xfxxxx例解解:利用函数表示与变量字母的无关的特性.,1xxt,11tx代入原方程得,111uux代入上式得设其中求令即即令即画线三式联立即例*.2421,lim(1)(1)(1)(1.)nnxxxxx当时求将分子、分母同乘以因子(1-x),则xxxxxxnn1)1()1)(1)(1)(1(lim242原式xxxxxnn1)1()1)(1)(1(lim2422xxxnnn1)1)(1(lim22xxnn11lim121.1x.)0lim,1(12nxxn时当例解!1):lim0nnnn证明222111:lim()112nnnnn2)证明:证明!121110lim0nnnnnnnnnnnn，!lim0nnnn22222111121nnnnnnnnn:证明22lim1,lim11nnnnnnn222111lim()112nnnnn例例.求.)321(lim1xxxx解:令xxxxf1)321()(xxx11)()(33231则)(xf3x133利用夹逼准则可知.3)(limxfxlim(1)(2)nnnnn原式12lim2nnnnn11112lim2l12im221nnnnnnnn1nn：本题对（+）进行有理化，注意做不到结果。例解11(2)3lim(2)3nnnnn22411limsinxxxxxx1121()333lim2()13nnn1322211141limsin110xxxxxxxxxxx（〈时同除）例例201sinlimsinxxxx2lim()1nnnn1211lim()lim()nnnnnnnen20000001sin1liml(0imlimsin0sinsin0)xxxxxxxxfxx20000001sin1liml(0imlimsin0sinsi0n)xxxxxxxxfxx0原式例例解例*.求解:xxeeexxxxsin12lim43401xxeexxxsin12lim410xxeexxxsin12lim4101原式=1(2000考研)xxeexxxsin12lim410121cos0lim(1)xxx22121cos0lim[(1)]xxxxx原式limlim[][lim]gxgxbfxfxa（）（）（利用（了）（））20lim1cosxxxe220lim22xxxee例解32sin!lim1nnnn201sin1lim1xxxxe32lim01nnnsin!1n0原式20sin2li12mxxxx无穷小量乘有界量仍为无穷小量注意书写！例例解32limlimsin!01nnnnn（）sin0limsintgxxxeetgxx201coslim(1)ln(1)xxxex2220022limlim22xxxxxxx（）-14sinsinsin00(1)sinlimlimsinsinxtgxxxxxeeetgxxtgxxtgxx（）1例例例(P70/4(6))解2011sinlim1sinxtgxxxxx021sinsinlim()(1sin11)xtgxxxtxgxx原式021sin(1)coslis2n2mixxxxx201coslimcosxxxx12202lim2xxx例.求下列极限：)sin1(sinlim)1(xxx解:xxsin1sin)1(21cos21sin2xxxx21cos)1(21sin2xxxx无穷小有界令1limx1xt0limt)1(sin)2(ttt0limttttsin)2(0limtttt)2(2xxsin12xxxsin112lim)2(解:0limxxxxcot110limxxxxcot)121(2e1解xxxxcot110lim)3(0limx12(1)1xxtgxxx0limx12(1)1xxx0limx12212(1)1xxxxx222lim2,??2xxaxbabxx22lim()420xxaxbab42ba2222242limlim2(2)(1)xxxaxbxaxaxxxx2(2)(2)(2)4lim2(2)(1)3xxxaxaxx2,8ab例解21lim()0,??1xxxx2211()(1)lim()lim()11xxxxxxxxx,11010,2()1lim(01xxxx（1-））例解(),1()??()31341ABxBxfxABfxxxx在x=1连续1()(4)4,()lim4313xABxBffxxx在x=1连续1lim[()]20xABxBAB2AB11()(2)limlim313313xxABxBBBxBxxxx1lim313xBxBxx1(1)(313)lim24313)xxxxBxxＢ（）-(2,4BA例解例.设函数在x=0连续,则a=,b=.解:20)cos1(lim)0(xxafx2a221~cos1xx)(lnlim)0(20xbfxblnbaln122e有无穷间断点及可去间断点解:为无穷间断点,)1)((lim0xaxbexx所以bexaxxx)1)((lim0ba101,0ba为可去间断点,)1(lim1xxbexx极限存在0)(lim1bexxeebxx1lim例.设函数试确定常数a及b.cos02()(0)?00xxxfxaaxaaxxx问时点连续00001(0),0lim(00)lim(00)(0)2xxfxfff点连续00cos1(00)lim,22xxfx00(00)limxaaxfx001lim()2xxxaaxa11212aa例解11()1xxfxe分析间断点及类型0110lim1xxxxe0x无穷间断点10111(10)lim11xxxxfe1x为跳跃间断点1011(10)lim01xxxfe例解例(P65/3(2))解()xfxtgx分析的间断点及类型xk1)0,0kx时0lim1xxtgx0x为第一类的可去间断点2)0klimxkxtgx(0)xkk为第二类的无穷间断点2xk2lim0xkxtgx2xk为第一类的可去间断点例(P75/11)解1,0()ln1xxexfxx分析间断点及类型（）,0x00x100(00)lim1,xxxfe00(00)limln10xfx（）0x为第一类的跳跃间断点1x110(10)lim0xxxfe110(10)limxxxfe1x为第二类的无穷间断点例.求的间断点,并判别其类型.解:)1)(1(sin)1(lim1xxxxxx1sin21x=–1为第一类可去间断点)(lim1xfxx=1为第二类无穷间断点,1)(lim0xfx,1)(lim0xfxx=0为第一类跳跃间断点1()lim1nxnxnefxe之连续性(,0)(00,)x连续为第一类的跳跃为间断点即区间11lim10111()lim001101010nxnnxnxnxnexeefxxex，=，，例解221()lim1nnnxfxxx之连续性例(P65/4)解,1()0,1,1xxfxxxx1x10(10)lim()1,xfx10(10)lim1xfx1x为第一类的跳跃间断点1x10(10)lim1,xfx10(10)lim1xfx1x