1.lim x→0，(tanx-sinx)/(sin2x)^3 2.lim x→0，sin(x-1)tan(x-1)/2(x-1)lnx 求高手详解啊。。。在线等
lim (tanx-sinx)/sin2x^3=\uff1f x\u8d8b\u4e8e0\u9519\u8bef\uff01\u5de6\u8fb9\u7b49\u4e8e1/4\uff0c\u53f3\u8fb9\u4e24\u4e2a\u6781\u9650\u503c\u90fd\u4e0d\u5b58\u5728\uff08\u90fd\u4e3a\u65e0\u7a77\u5927\uff09
\u505a\u6cd5\u5982\u4e0b\uff1a
\u7531\u4e8ex\u8d8b\u4e8e0\uff0c\u53ef\u4ee5\u7b49\u53f7\u5de6\u8fb9\u770b\u51fa\u5206\u5b50\u90fd\u8d8b\u4e8e0\uff0c\u53ef\u4ee5\u4e0a\u4e0b\u540c\u65f6\u6c42\u5bfc\uff0c
\uff081/cos\u5e73\u65b9x-cosx\uff09/6x\u5e73\u65b9cos2x\u76843\u6b21\u65b9\u3002\u5206\u5b50\u5206\u6bcd\u540c\u65f6\u4e58\u4ee5cos\u5e73\u65b9x \u5e76\u4e14\u5206\u6bcd\u4e2d\u5c06x=0\u628a\u6240\u6709\u7684cosx\u548ccos2x\u90fd\u53ef\u4ee5\u6362\u62101(\u53ef\u4ee5\u7406\u89e3\u4e3a\u5c06\u5206\u6bcd\u4e2d\u542b\u6709cosx\u548ccos2x\u7684\u9879\u63d0\u51fa\u6765\uff0c\u53d1\u73b0x\u8d8b\u4e8e0\u6781\u9650\u5b58\u5728\uff0c\u6240\u4ee5\u53ef\u4ee5\u66ff\u6362\u6389)
\u5f97\u5230 (1-cos\u76843\u6b21\u65b9)/6x\u5e73\u65b9 \uff0c\u7136\u540e\u53ef\u4ee5\u7ee7\u7eed\u4e0a\u4e0b\u540c\u65f6\u6c42\u5bfc\uff0c
3cos\u5e73\u65b9xsinx/12x , \u53ef\u4ee5\u7ee7\u7eed\u628acosx\u6362\u62101\u3002 \u7ee7\u7eed\u6c42\u5bfc\uff0c\u5f97\u5230
3cosx/12 \uff0c\u4e5f\u5c31\u662f\u5728x\u8d8b\u4e8e0\u7684\u65f6\u5019\uff0c\u6781\u9650\u503c\u4e3a1/4\u3002
\u800c\u53f3\u8fb9\u4e24\u4e2a\u6781\u9650\u90fd\u4e0d\u5b58\u5728\uff0c\u65e0\u6cd5\u8ba1\u7b97\u3002
\u65e0\u7a77\u5c0f\u7684\u5b9a\u4e49\u662f\u4ee5\u6781\u9650\u7684\u5f62\u5f0f\u6765\u5b9a\u4e49\u7684\uff0c\u5f53x\u2192x0\u65f6(\u6216x\u2192\u221e)\u65f6\uff0climf(x)=0\uff0c\u5219\u79f0\u51fd\u6570f(x)\u5f53x\u2192x0\u65f6(\u6216x\u2192\u221e)\u65f6\u4e3a\u65e0\u7a77\u5c0f\u3002
\u5f53lim\u03b2\u03b1=1\uff0c\u5c31\u8bf4\u03b2\u4e0e\u03b1\u662f\u7b49\u4ef7\u65e0\u7a77\u5c0f\u3002
\u5e38\u89c1\u6027\u8d28\u6709\uff1a
\u8bbe\u03b1,\u03b1\u2032,\u03b2,\u03b2\u2032,\u03b3 \u7b49\u5747\u4e3a\u540c\u4e00\u81ea\u53d8\u91cf\u53d8\u5316\u8fc7\u7a0b\u4e2d\u7684\u65e0\u7a77\u5c0f\uff0c \u2460 \u82e5\u03b1\uff5e\u03b1\u2032,\u03b2\uff5e\u03b2\u2032\uff0c \u4e14lim\u03b1\u2032\u03b2\u2032\u5b58\u5728\uff0c\u5219lim\u03b1\u03b2=lim\u03b1\u2032\u03b2\u2032\u2461 \u82e5\u03b1\uff5e\u03b2\uff0c\u03b2\uff5e\u03b3\uff0c\u5219\u03b1\uff5e\u03b3
\u6027\u8d28\u2460\u8868\u660e\u7b49\u4ef7\u65e0\u7a77\u5c0f\u91cf\u7684\u5546\u7684\u6781\u9650\u6c42\u6cd5\u3002\u6027\u8d28\u2461\u8868\u660e\u7b49\u4ef7\u65e0\u7a77\u5c0f\u7684\u4f20\u9012\u6027\u82e5\u80fd\u8fd0\u7528\u6781\u9650\u7684\u8fd0\u7b97\u6cd5\u5219\uff0c\u53ef\u7ee7\u7eed\u62d3\u5c55\u51fa\u4e0b\u5217\u7ed3\u8bba\uff1a
\u2462 \u82e5\u03b1\uff5e\u03b1\u2032,\u03b2\uff5e\u03b2\u2032\uff0c \u4e14lim\u03b2\u03b1=c(\u2260-1)\uff0c\u5219\u03b1+\u03b2\uff5e\u03b1\u2032+\u03b2\u2032
\u8bc1\u660e\uff1a\u2235 lim\u03b1+\u03b2\u03b1\u2032+\u03b2\u2032=lim1+\u03b2\u03b1\u03b1\u2032\u03b1+\u03b2\u2032\u03b1\u2032=lim1+c1+\u03b1\u03b1\u2032\u00b7\u03b2\u03b1\u00b7\u03b2\u2032\u03b2
=lim1+c1+c=1 \u2234 \u03b1+\u03b2\uff5e\u03b1\u2032+\u03b2\u2032
\u800c\u5b66\u751f\u5219\u5f80\u5f80\u5728\u6027\u8d28(3)\u7684\u5e94\u7528\u4e0a\u5ffd\u7565\u4e86\u201clim\u03b2\u03b1=c(\u2260-1)\u201d\u8fd9\u4e2a\u6761\u4ef6\uff0c\u5343\u7bc7\u4e00\u5f8b\u8ba4\u4e3a\u201c\u03b1\uff5e\u03b1\u2032,\u03b2\uff5e\u03b2\u2032\uff0c\u5219\u6709\u03b1+\u03b2\uff5e\u03b1\u2032+\u03b2\u2032
\u2463 \u82e5\u03b1\uff5e\u03b1\u2032,\u03b2\uff5e\u03b2\u2032\uff0c \u4e14limA\u03b1\u2032\u00b1B\u03b2\u2032C\u03b1\u2032\u00b1D\u03b2\u2032\u5b58\u5728\uff0c\u5219\u5f53A\u03b1\u2032\u00b1B\u03b2\u2032C\u03b1\u2032\u00b1D\u03b2\u2032\u22600\u4e14 limA\u03b1\u00b1B\u03b2C\u03b1\u00b1D\u03b2\u5b58\u5728\uff0c\u6709limA\u03b1\u00b1B\u03b2C\u03b1\u00b1D\u03b2=limA\u03b1\u2032\u00b1B\u03b2\u2032C\u03b1\u2032\u00b1D\u03b2\u2032
\u503c\u5f97\u6ce8\u610f\u7684\u662f\uff0c\u7b49\u4ef7\u65e0\u7a77\u5c0f\u53ea\u6709\u5728\u6781\u9650\u5b58\u5728\u4e14\u53ef\u6c42\u7684\u60c5\u51b5\u4e0b\u66ff\u6362\uff0c\u5426\u5219\u53ea\u80fd\u4e00\u6b65\u4e00\u6b65\u7528\u6d1b\u5fc5\u8fbe\u6cd5\u5219\u4e86\u3002\u8fd9\u4e2a\u6cd5\u5219\u8fd8\u662f\u4e0d\u9ebb\u70e6\u7684\uff0c\u6bd5\u7adf\u5fae\u5206\u8fd8\u662f\u5f88\u7b80\u5355\u7684\u8fd0\u7b97\u3002\u3002\u3002
\u3002\u3002\u3002\u5206\u6bcd\u662f\uff08sin2x\uff09\u76843\u6b21\u65b9\uff0c\u8fd8\u662fsin(2*x\u76843\u6b21\u65b9) \uff1f
\u90a3\u6211\u5c31\u4e0d\u7528\u6d1b\u5fc5\u8fbe\u6cd5\u5219\u4e86\u5475\u5475~,\u7528\u5b9a\u7406lim[x\u21920] sinx/x=1
lim[x\u21920] (tanx-sinx)/x³
=lim[x\u21920] (sinx/cosx-sinx)/x³
=lim[x\u21920] (sinx-sinxcosx)/(x³cosx)
=lim[x\u21920] sinx(1-cosx)/(x³cosx)
=lim[x\u21920] sin³x(1-cosx)/(x³sin²xcosx)
=lim[x\u21920] (sinx/x)³\u00b7(1-cosx)/(sin²xcosx)
=lim[x\u21920] (sinx/x)³\u00b7(1-cosx)/[(1-cos²x)cosx]
=lim[x\u21920] (sinx/x)³\u00b7(1-cosx)/[(1+cosx)(1-cosx)cosx]
=lim[x\u21920] (sinx/x)³\u00b71/[(1+cosx)cosx]
=1\u00b71/(1+1)
=1/2
应用罗比达法则，分子分母同时求导
上式=lim x→0，(1/(cosx)^2-cosx)/[3*（sin2x)^2*cos2x*2]
=1/6lim x→0(1-(cosx)^3)/(sin2x)^2
再次应用罗比达法则，分子分母同时求导
上式=1/6lim x→0（3(cosx)^2*sinx)/2sin2x*cos2x*2)=1/8lim x→0(sinx/sin2x)=1/16
2.lim x→0，sin(x-1)tan(x-1)/2(x-1)lnx
=-(sin1*tan1)/2lim x→0，1/lnx=0
第一题可以不用求导得罗必塔法则，可以直接求出
(tanx-sinx)/(sin2x)^3
=sinx(1-cosx)/[cosx(2sinxcosx)^3]
=sinx(1-cosx)/[8(sinx)^3(cosx)^4]
=(1-cosx)/[8(sinx)^2(cosx)^4]
=(1-cosx)/{8[1-(cosx)^2](cosx)^4}
=(1-cosx)/8[(1-cosx)(1+cosx)(cosx)^4]
=1/8 * 1/[(1+cosx)(cosx)^4]
当x->0, cosx->1
所以原极限式=1/16
第二题
sin(x-1)tan(x-1)/2(x-1)lnx
=-(sin1*tan1)/2
扩展阅读：lim 1+x ... lim tanx tan2x ... 极限公式大全24个 ... lim0 0 ... lim 1 x tan二分之派x ... 计算极限tan x lim ... lim x0 ... 极限limx0sin2x除以x ... limx ...
