同济大学高等数学一、求下列极限1、sin()limxxx→−−22111 ；解一：( ) ( )12sin1cos1lim02xxxx→−−==原式解二：( )( )11sin1sin1limlim011xxxxxx→→−−==−+原式2、limsinxxx→2203解一：00021311limlimlim6sin3cos39sin3cos39xxxxxxxxx→→→==⋅=原式解二：sin3 ~30021limlim6sin3 cos39 cos39xxxxxxxxxx→→===原式3、20tan2limsin3xxxx→解：()2tan2~2,sin3~32022lim93x xx xxxx→=原式=4、0limln(1)xxx→+解一：()001limlim1111xxxx→→==+=+原式解二：()1011lim1lnln1xxex→===+原 式5、2limxxxx→∞−⎛⎞⎜⎟⎝⎠解一：( )2222lim1xxex− ⋅−−→∞⎛⎞=−=⎜⎟⎝⎠原式解二：( )1211ln 2 ln22limlimln2lim22lim xxxxxxxxxxxxxxxeeeee−−→∞→∞→∞−− −−−−→∞−−−=====原式6、()111lim3 2xxx −→−解一：()()112220lim1 2txttte=−− ⋅−−→=−=令原式解二：1(2)221122221lim[1(22)] {lim[1(22)]}xxxxxxe−−→−−−→=+ −=+ −=i原式7、30sinlimxxxx→−解：2001 cossin1limlim366xxxxxx→→−===原式8、111limln1xxx→⎛⎞−⎜⎟−⎝⎠解：11111 1ln11limlimlim1( 1)lnln1ln11 limln 1 12xxxxx xxxxxxx x xxxx→→→→−− +−===−−+ −+−==−+ +原式9、1 2lim 22nn nn→∞+ + +⎛⎞−⎜⎟+⎝⎠⋯解：( )()221122limlim22221 lim4 22nnnnnnn nn nnnnn→∞→∞→∞⎛⎞+⎜⎟+ −−=−=⎜⎟++⎜⎟⎝⎠−==−+原式10、32090sinlimxxtdtx→∫解：26686003 sin 1sin 1limlim933xxxxxxx→→===原式11、02arctanlim1xxtdtx→+∞+∫。( )12222arctan1limlimarctan111221limarctanlim1122xxxxxx xxxxxππ→+∞→+∞−→+∞→+∞==⋅++⋅=⋅+ = ⋅=解：原式二、求下列导数或微分1、设4tan1y xxx=−+ ，求dy解一：dyydx′=()324tan secxx xxdx=−−解二：()34tantandy xdxxdxxd x=−+()32324tansec4tan secxdxxdxxxdxxx xxdx=−− ⋅=−−2、设21 cosxyx= +，求 y′() ()()()()222ln2 1 cos 2sin1 cos2 sincosln2 ln2 1 cosxxxxxyxxxx⋅ +−−′=++⋅+=+解:3、设51logsinyx=，求 y′( )221cos111cos11ln5sinln5sinxyxxxxx−′=⋅⋅−=−⋅⋅解 4、设2ln( 1)yxx=++，求 y′′( )( )( )( )122222212233222211112211111111212yxxxxxxxxxxyxxxx−−−−⎛⎞′/=⋅ ++⋅⎜⎟/++⎝⎠++=⋅+++= +′′=−+⋅ =−+解：5、设lnx yy=...
