某道三角函数求极限题的变换方式求解: lim(1-cosx+sinx)=lim(sinx) (x趋向于0)这步变换是怎么完成的?
\u6c42\u6781\u9650lim \uff08sinx/x\uff09^(1/1-cosx)\uff0cx\u8d8b\u5411\u4e8e0\u56e0\u4e3asinx/x \u53ea\u662f\u53d6x\u8d8b\u5411\u4e8e0\u65f6\u7684\u6781\u9650\u503c\u4e3a1\uff0c\u800c\u6307\u6570\u7684\u6781\u9650\u662f\u221e\u3002\u6781\u96501\u7684\u221e\u6b21\u65b9\u662f\u4e0d\u5b9a\u5f0f\u3002
lim \uff08sinx/x\uff09^(1/1-cosx)
=e^ lim (1/1-cosx)\u00b7ln(sinx/x)
=e^ lim ( 1/(x²/2) )\u00b7ln(1+ sinx/x -1)\u3010\u7b49\u4ef7\u4ee3\u6362\uff1a1-cosx\uff5ex²/2\u3011
=e^ lim ( 2/x² )\u00b7(sinx/x - 1)\u3010\u7b49\u4ef7\u4ee3\u6362\uff1au\u21920\u65f6ln(1+u)\uff5eu\u3011
=e^ 2lim ( (sinx - x) / x³ )
=e^ 2lim ( (cosx - 1) / (3x²) )\u3010\u6d1b\u6bd4\u8fbe\u6cd5\u5219\u3011
=e^ 2lim ( (-x²/2) / (3x²) )\u3010\u7b49\u4ef7\u4ee3\u6362\uff1a1-cosx\uff5ex²/2\u3011
=e^(-1/3)
x\u00b0=x/360*2\u03c0\u5f27\u5ea6
\u56e0\u6b64
lim sinx\u00b0/x (x->0)
=2\u03c0/360
=\u03c0/180
lim(1-cosx+sinx)=lim(1-cosx)+limsinx=0+lim(sinx)=lim(sinx)
运用等价无穷小的性质,X趋向与0,SINX~X,COSX~1,1-1等于0
不能 我猜的。
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