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习题191求函数633)(223xxxxxxf的连续区间并求极限)(lim0xfx)(lim3xfx及)(lim2xfx解)2)(3()1)(1)(3(633)(223xxxxxxxxxxxf函数在()内除点x2和x3外是连续的所以函数f(x)的连续区间为(3)、(32)、(2)在函数的连续点x0处21)0()(lim0fxfx在函数的间断点x2和x3处)2)(3()1)(1)(3(lim)(lim22xxxxxxfxx582)1)(1(lim)(lim33xxxxfxx2设函数f(x)与g(x)在点x0连续证明函数(x)max{f(x)g(x)}(x)min{f(x)g(x)}在点x0也连续证明已知)()(lim00xfxfxx)()(lim00xgxgxx可以验证]|)()(|)()([21)(xgxfxgxfx]|)()(|)()([21)(xgxfxgxfx因此]|)()(|)()([21)(00000xgxfxgxfx]|)()(|)()([21)(00000xgxfxgxfx因为]|)()(|)()([21lim)(lim00xgxfxgxfxxxxx]|)(lim)(lim|)(lim)(lim[210000xgxfxgxfxxxxxxxx]|)()(|)()([210000xgxfxgxf(x0)所以(x)在点x0也连续同理可证明(x)在点x0也连续3求下列极限(1)52lim20xxx(2)34)2(sinlimxx(3))2cos2ln(lim6xx(4)xxx11lim0(5)145lim1xxxx(6)axaxaxsinsinlim(7))(lim22xxxxx解(1)因为函数52)(2xxxf是初等函数f(x)在点x0有定义所以55020)0(52lim220fxxx(2)因为函数f(x)(sin2x)3是初等函数f(x)在点4x有定义所以1)42(sin)4()2(sinlim334fxx(3)因为函数f(x)ln(2cos2x)是初等函数f(x)在点6x有定义所以0)62cos2ln()6()2cos2ln(lim6fxx(4))11(lim)11()11)(11(lim11lim000xxxxxxxxxxxx211101111lim0xx(5))45)(1()45)(45(lim145lim11xxxxxxxxxxxx)45)(1(44lim1xxxxx214154454lim1xxx(6)axaxaxaxaxaxax2sin2cos2limsinsinlimaaaaxaxaxaxaxcos12cos22sinlim2coslim(7))())((lim)(lim22222222xxxxxxxxxxxxxxxxxx1)1111(2lim)(2lim22xxxxxxxxx4求下列极限(1)xxe1lim(2)xxxsinlnlim0(3)2)11(limxxx(4)xxx2cot20)tan31(lim(5)21)63(limxxxx(6)xxxxxx20sin1sin1tan1lim解(1)1lim01lim1eeexxxx(2)01ln)sinlimln(sinlnlim00xxxxxx(3)eexxxxxx21212)11(lim)11(lim(4)33tan3120cot2022)tan31(lim)tan31(limexxxxxx(5)21633621)631()63(xxxxxxx因为exxx36)631(lim232163limxxx所以2321)63(limexxxx(6))sin1tan1)(1sin1()1sin1)(sin1tan1(limsin1sin1tan1lim22020xxxxxxxxxxxxxxxxxxxxxxxxxxx220220sin2sin2tanlim)sin1tan1(sin)1sin1)(sin(tanlim21)2(2lim320xxxx5设函数00)(xxaxexfx应当如何选择数a使得f(x)成为在()内的连续函数？解要使函数f(x)在()内连续只须f(x)在x0处连续即只须afxfxfxx)0()(lim)(lim00因为1lim)(lim00xxxexfaxaxfxx)(lim)(lim00所以只须取a1