习题19x33x2x31.求函数f(x)的连续区间,并求极限limf(x),limf(x)及limf(x).x2x6x0x3x2x33x2x3(x3)(x1)(x1)解f(x),函数在(,)内除点x2和x3外是连续x2x6(x3)(x2)的,所以函数f(x)的连续区间为(,3)、(3,2)、(2,).1在函数的连续点x0处,limf(x)f(0).x02在函数的间断点x2和x3处,(x3)(x1)(x1)(x1)(x1)8limf(x)lim,limf(x)lim.x2x2(x3)(x2)x3x3x252.设函数f(x)与g(x)在点x连续,证明函数0(x)max{f(x),g(x)},(x)min{f(x),g(x)}在点x也连续.0证明已知limf(x)f(x),limg(x)g(x).00xxxx00可以验证1(x)[f(x)g(x)|f(x)g(x)|],21(x)[f(x)g(x)|f(x)g(x)|].21因此(x)[f(x)g(x)|f(x)g(x)|],0200001(x)[f(x)g(x)|f(x)g(x)|].020000因为1lim(x)lim[f(x)g(x)|f(x)g(x)|]xxxx2001[limf(x)limg(x)|limf(x)limg(x)|]2xxxxxxxx00001[f(x)g(x)|f(x)g(x)|](x),200000所以(x)在点x也连续.0同理可证明(x)在点x也连续.03.求下列极限:(1)limx22x5;x0(2)lim(sin2x)3;x4(3)limln(2cos2x)x6x11(4)lim;x0x5x4x(5)lim;x1x1sinxsina(6)lim;xaxa(7)lim(x2xx2x).x解(1)因为函数f(x)x22x5是初等函数,f(x)在点x0有定义,所以limx22x5f(0)022055.x0(2)因为函数f(x)(sinx2)3是初等函数,f(x)在点x有定义,所以4lim(sin2x)3f()(sin2)31.x444(3)因为函数f(x)ln(2cos2x)是初等函数,f(x)在点x有定义,所以6limln(2cos2x)f()ln(2cos2)0.x666x11(x11)(x11)x111(4)limlimlimlim.x0xx0x(x11)x0x(x11)x0x1101125x4x(5x4x)(5x4x)4x4(5)limlimlimx1x1x1(x1)(5x4x)x1(x1)(5x4x)44lim2.x15x4x5141xaxa2cossinsinxsina22(6)limlimxaxaxaxaxasinxaaalimcoslim2cos1cosa.xa2xaxa22(x2xx2x)(x2xx2x)(7)lim(x2xx2x)limxx(x2xx2x)2x2limlim1.x22x11(xxxx)(11)xx4.求下列极限:1(1)limex;xsinx(2)limln;x0x1x(3)lim(1)2;xx(4)lim(13tan2x)cot2x;x03xx1(5)lim()2;x6x1tanx1sinx(6)lim.x0x1sin2xx11lim解(1)limexexxe01.xsinxsinx(2)limlnln(lim)ln10.x0xx0xx1111x2(3)lim(1)2lim(1)2ee.xxxx1(4)lim(13tan2x)cot2xlim(13tan2x)3tan2x3e3.x0x03xx136x3x1(5)(