求极限，当x趋于0时，tansinx-sintanx除以tanx-sinx的极限 下列这道极限怎么解？lim（x趋于0）（tantanx-si...
\u5f53x\u8d8b\u8fd1\u4e0e0\u65f6,\u6c42(tanx-sinx)\u7684\u6781\u9650\u63d0\u51fatan\uff0c\u7136\u540e\u53d8\u6210tanx(1-cosx)\uff0cx\u8d8b\u4e8e0\uff0ctan=0\uff0c1-cos x~x^2/2=0
sinx=x-x^3/6+o(x^3)
sin(sinx)=sinx-(sinx)^3/6+o(sinx)
=x-x^3/6+o(x^3)-(x-x^3/6+o(x^3))^3/6 +o(sinx)
=x-x^3/3+o(x^3)
tanx=x+x^3/3+o(x^3)
tan(tanx)=x+x^3/3+(x+x^3+o(x^3))^3/3+o(x^3)
=x+2/3 x^3+o(x^3)
tanx-sinx=x^3/2+o(x^3)
\u6240\u4ee5\u6c42\u6781\u9650
=lim(x-->0)(x^3+o(x^3))/(x^3/2 x^3-o(x^3))
=2
\u4e0d\u5b9a\u79ef\u5206\u7684\u516c\u5f0f\uff1a
1\u3001\u222b a dx = ax + C\uff0ca\u548cC\u90fd\u662f\u5e38\u6570
2\u3001\u222b x^a dx = [x^(a + 1)]/(a + 1) + C\uff0c\u5176\u4e2da\u4e3a\u5e38\u6570\u4e14 a \u2260 -1
3\u3001\u222b 1/x dx = ln|x| + C
4\u3001\u222b a^x dx = (1/lna)a^x + C\uff0c\u5176\u4e2da > 0 \u4e14 a \u2260 1
5\u3001\u222b e^x dx = e^x + C
6\u3001\u222b cosx dx = sinx + C
7\u3001\u222b sinx dx = - cosx + C
8\u3001\u222b cotx dx = ln|sinx| + C = - ln|cscx| + C
用罗比达法则
= （（secsin x)**2 *cos x - costanx* (sec x）**2 ）/((secx)**2+cos x)
=(0-1)/(1+1） =- 1/2
0
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