lim(x→0){[ln(1+(sinx)^2)-6*((2-cosx)^(1/3)-1)]/x^4} 用泰勒公式做lim(x→0){[ln(1+(sinx)^2)-6*((2-cosx)^(1/3)-1)]/x^4}用泰勒公式做,答案是-7/12我是这么做的:ln(1+(sinx)^2)=x^2-(5/6)x^4+o(x^4)(2-cosx)^(1/3)=1+(1/6)x^2-(1/9*24)x^4
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![lim(x→0){[ln(1+(sinx)^2)-6*((2-cosx)^(1/3)-1)]/x^4} 用泰勒公式做lim(x→0){[ln(1+(sinx)^2)-6*((2-cosx)^(1/3)-1)]/x^4}用泰勒公式做,答案是-7/12我是这么做的:ln(1+(sinx)^2)=x^2-(5/6)x^4+o(x^4)(2-cosx)^(1/3)=1+(1/6)x^2-(1/9*24)x^4](/uploads/image/z/5348799-63-9.jpg?t=lim%28x%E2%86%920%29%7B%5Bln%281%2B%28sinx%29%5E2%29-6%2A%28%282-cosx%29%5E%281%2F3%29-1%29%5D%2Fx%5E4%7D+%E7%94%A8%E6%B3%B0%E5%8B%92%E5%85%AC%E5%BC%8F%E5%81%9Alim%28x%E2%86%920%29%7B%5Bln%281%2B%28sinx%29%5E2%29-6%2A%28%282-cosx%29%5E%281%2F3%29-1%29%5D%2Fx%5E4%7D%E7%94%A8%E6%B3%B0%E5%8B%92%E5%85%AC%E5%BC%8F%E5%81%9A%2C%E7%AD%94%E6%A1%88%E6%98%AF-7%2F12%E6%88%91%E6%98%AF%E8%BF%99%E4%B9%88%E5%81%9A%E7%9A%84%EF%BC%9Aln%281%2B%28sinx%29%5E2%29%3Dx%5E2-%285%2F6%29x%5E4%2Bo%28x%5E4%29%282-cosx%29%5E%281%2F3%29%3D1%2B%281%2F6%29x%5E2-%281%2F9%2A24%29x%5E4)
lim(x→0){[ln(1+(sinx)^2)-6*((2-cosx)^(1/3)-1)]/x^4} 用泰勒公式做lim(x→0){[ln(1+(sinx)^2)-6*((2-cosx)^(1/3)-1)]/x^4}用泰勒公式做,答案是-7/12我是这么做的:ln(1+(sinx)^2)=x^2-(5/6)x^4+o(x^4)(2-cosx)^(1/3)=1+(1/6)x^2-(1/9*24)x^4
lim(x→0){[ln(1+(sinx)^2)-6*((2-cosx)^(1/3)-1)]/x^4} 用泰勒公式做
lim(x→0){[ln(1+(sinx)^2)-6*((2-cosx)^(1/3)-1)]/x^4}
用泰勒公式做,答案是-7/12
我是这么做的:
ln(1+(sinx)^2)=x^2-(5/6)x^4+o(x^4)
(2-cosx)^(1/3)=1+(1/6)x^2-(1/9*24)x^4+o(x^4)
lim(x→0){[ln(1+(sinx)^2)-6*((2-cosx)^(1/3)-1)]/x^4}=[x^2-(5/6)x^4-6*(1+(1/6)x^2-(1/9*24)x^4)]/x^4=-29/36
为什么不对呢?
lim(x→0){[ln(1+(sinx)^2)-6*((2-cosx)^(1/3)-1)]/x^4} 用泰勒公式做lim(x→0){[ln(1+(sinx)^2)-6*((2-cosx)^(1/3)-1)]/x^4}用泰勒公式做,答案是-7/12我是这么做的:ln(1+(sinx)^2)=x^2-(5/6)x^4+o(x^4)(2-cosx)^(1/3)=1+(1/6)x^2-(1/9*24)x^4
问题在这里,(2-cosx)^(1/3)=1+(1/6)x^2-(1/9*24)x^4+o(x^4)
x^4的系数错了
应该是(2-cosx)^(1/3)=(1-x^2/2+x^4/24+o(x^4))^(1/3)=1+(1/6)x^2-(1/24)x^4+o(x^4)