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文档来源为:从网络收集整理.word版本可编辑.欢迎下载支持.习题1-11.计算下列极限xaax(1)lim,a0;xaxaxaaaaaxaxa解:原式lim[]=(a)|(x)|xaxaxaxaxaaa1a=alnaaa=a(lna1)sinxsina(2)lim;xasin(xa)sinxsina解:原式lim(sinx)'xacosaxaxa2n1(3)limn(a2),a0;nna1na12x22解:原式lim()[(a)']lnanna1/nx01(4)limn[(1)p1],p0;nn(11)p1ppp1解:原式limn1(x)|x1pxx1pnn1010(1tanx)(1sinx)(5)lim;x0sinx1010(1tanx)1(1sinx)1解:原式limlimx0tanxx0sinx99=10(1t)|10(1t)|20t0t0mx1(6)lim,m,n为正整数;x1nx11(xm)'mx1x1x1n解:原式limx1x1nx11m(xn)'x1f(xh)2f(x)f(xh)2.设f(x)在x处二阶可导,计算lim000.02h0hf(xh)f(xh)f(xh)f(x)f(x)f(xh)解:原式lim00lim0000h02hh02h1f(x)lnxlna3.设a0,f(a)0,f(a)存在,计算lim[].xaf(a)1lnf(x)lnf(a)解:lim[f(x)]lnxlnalimelnxlnaxaf(a)xa习题1-21文档来源为:从网络收集整理.word版本可编辑.文档来源为:从网络收集整理.word版本可编辑.欢迎下载支持.1.求下列极限(1)lim(sinx1sinx1);x1解:原式limcos[(x1)(x1)]0,其中在x1与x1之间x2cos(sinx)cosx(2)lim;4x0sinx解:原