已知limx趋近于[6sinx-(tanx)f(x)]/x^3=0 求limx趋近于0[6-f(x)]/x^2[sin6xcosx-sinxf(x)]/x^3cosx={sin6xcosx-6sinx+[6-f(x)]sinx}/x^3cosx=(sin6xcosx-6sinx)/x^3cosx+要求极限=0 得到要求的极限=(6sinx-sin6xcosx)/x^3cosx=0但是答案上(6si
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![已知limx趋近于[6sinx-(tanx)f(x)]/x^3=0 求limx趋近于0[6-f(x)]/x^2[sin6xcosx-sinxf(x)]/x^3cosx={sin6xcosx-6sinx+[6-f(x)]sinx}/x^3cosx=(sin6xcosx-6sinx)/x^3cosx+要求极限=0 得到要求的极限=(6sinx-sin6xcosx)/x^3cosx=0但是答案上(6si](/uploads/image/z/12966826-58-6.jpg?t=%E5%B7%B2%E7%9F%A5limx%E8%B6%8B%E8%BF%91%E4%BA%8E%5B6sinx-%28tanx%29f%28x%29%5D%2Fx%5E3%3D0+%E6%B1%82limx%E8%B6%8B%E8%BF%91%E4%BA%8E0%5B6-f%28x%29%5D%2Fx%5E2%5Bsin6xcosx-sinxf%28x%29%5D%2Fx%5E3cosx%3D%7Bsin6xcosx-6sinx%2B%5B6-f%28x%29%5Dsinx%7D%2Fx%5E3cosx%3D%28sin6xcosx-6sinx%29%2Fx%5E3cosx%2B%E8%A6%81%E6%B1%82%E6%9E%81%E9%99%90%3D0+%E5%BE%97%E5%88%B0%E8%A6%81%E6%B1%82%E7%9A%84%E6%9E%81%E9%99%90%3D%286sinx-sin6xcosx%29%2Fx%5E3cosx%3D0%E4%BD%86%E6%98%AF%E7%AD%94%E6%A1%88%E4%B8%8A%286si)
已知limx趋近于[6sinx-(tanx)f(x)]/x^3=0 求limx趋近于0[6-f(x)]/x^2[sin6xcosx-sinxf(x)]/x^3cosx={sin6xcosx-6sinx+[6-f(x)]sinx}/x^3cosx=(sin6xcosx-6sinx)/x^3cosx+要求极限=0 得到要求的极限=(6sinx-sin6xcosx)/x^3cosx=0但是答案上(6si
已知limx趋近于[6sinx-(tanx)f(x)]/x^3=0 求limx趋近于0[6-f(x)]/x^2
[sin6xcosx-sinxf(x)]/x^3cosx={sin6xcosx-6sinx+[6-f(x)]sinx}/x^3cosx=(sin6xcosx-6sinx)/x^3cosx+要求极限=0 得到要求的极限=(6sinx-sin6xcosx)/x^3cosx=0
但是答案上(6sinx-sin6xcosx)/x^3=0①将分母处的cosx=1省去,得到6sinx(1-cosx)/x^3+cosx(6sinx+sin6x)/x^3=0.5*6+(6sinx+sin6x)/x^3=0.5*6+35=38
我想知道这里的COSX为什么在分母处直接当作1省略,而且答案上的第二式的分子也可以将COSX当作1省略,而如果不省略,则很明显为0,而且如果将①式中分子第二项的cosx也做为1省略的话,由洛必达也很容易得出为35,所以就特别纠结。35,37,38四个选项。
PS:题目中的第一个打错了 是SIN6X
已知limx趋近于[6sinx-(tanx)f(x)]/x^3=0 求limx趋近于0[6-f(x)]/x^2[sin6xcosx-sinxf(x)]/x^3cosx={sin6xcosx-6sinx+[6-f(x)]sinx}/x^3cosx=(sin6xcosx-6sinx)/x^3cosx+要求极限=0 得到要求的极限=(6sinx-sin6xcosx)/x^3cosx=0但是答案上(6si
x趋近0时,sinx与x、tanx与x都是等价无穷小
即,lim(x->0)(sinx/x)=1
lim(x->0)(tanx/x)=1
limx趋近于0[6sinx-(tanx)f(x)]/x³
=limx趋近于0[6-f(x)]/x²
=0
所以,limx趋近于0[6-f(x)]/x²=0