1第一章函数与极限一、要求:函数定义域,奇偶性判定,反函数,复合函数分解,渐近线,求极限,间断点类型判定,分段函数分段点连续性判定及求未知参数,零点定理应用.二、练习:1.函数2112xxy的定义域;答:2x且1x;2.函数2lnsinyx是由:复合而成的;答:2,ln,,sinyuuvvwwx;3.设,1122xxxxf则()fx=;答:22x;4.已知2110fxxxx,则fx;答:2211111xfxxxxx0x;5.11lim1nxxx=,答:n;32sin!lim1nnnn=;答:0;6.当a时,函数,0,0xexfxaxx在(,)上连续;答:1a;7.设(3)(3)fxxx,则(3)fx(B);A.(3)xx,B.6(3)xx,C.6(3)xx,D.(3)(3)xx;8.1limsinnnn(B);A.0,B.1,C.,D.;9.1x是函数221()32xfxxx的(A);A.可去间断点,B.跳跃间断点,C.第二类间断点,D.连续点;10.|sin|()cosxfxxxe是(A);A.奇函数,B.周期函数,C.有界函数,D.单调函数;11.下列正确的是(A)A.1limsin0xxx,B.1limsin0xxx,C.01limsin1xxx,D.11limsin1xxx;12.1x是函数1,13,1xxfxxx的(D)2A、连续点B、可去间断点C、第二类间断点D、跳跃间断点13.函数221xxy的反函数为(A)A.2log0,11xyxx,B.2log1yxy,C.2log1xyx,D.ln1xyx14.计算221lim1xxxx;2lim1xxxx;(3)30tansinlimsin2xxxx;(4)21/30(1)1limcos1xxx;(5)231lim3cosxxxxx;(6)4213lim22xxx;(7)20lnln2lnlimxaxaxax;(8)22sin1limarctan1xxxxx;(9)011limln1xxxx;(10)221limarcsin11xxxx.解:22121111limlimlim11111111xxxxxxxxxeexxx;12limlim11xxxxxexx;(3)23330001tan1costansin12limlimlimsin28816xxxxxxxxxxxx;(4)221/30021(1)123limlim1cos132xxxxxx;(5)223311lim0,23cos4,lim3cos0xxxxxxxxxx;(6)444242222221322limlimlim3224213213xxxxxxxxxxx;(7)2222222222000ln1lnln2ln11limlimlimln1axxxxxaxaxaaxxxaaa;3或原式2220ln1limxxax222201limxxaxa(8)2222sin12121limarctanlimsinlimlimsin200111xxxxxxxxxxxxxxxx;或原式=222sin12sin1limarctanlimlimarctan11xxxxxxxxxxx=0(9)1100011111limlnlimln1ln1limln1ln11122xxxxxxxxxxxxx;或00000111111112112112limlnlimlnlimln1limlim1111122221111xxxxxxxxxxxxxxxxxx(10)22221212limarcsinlimlim0111111xxxxxxxxxx.15.已知21lim31xxaxbx,求,ab的值;解:设21xaxbxxc,则11lim13,21xxxcccx,所以11,2acbc.16.已知232lim43xxxkx,求k的值.解:2332lim4,lim303xxxxkxx,23lim230,3xxxkkk.17.证明方程3320xx在区间1,1内至少有一个根.证明设332fxxx,则fx在闭区间1,1上连续,又113220,113260,ff由零点定理,至少存在一点1,1,使0f;即3320f,即方程3320xx在区间1,1内至少有一个根.