作业帮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+
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/28 23:08:41
![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/3859464-48-4.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%2B)
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+
加减不可以用无穷小量代替无穷小量,你可以尝试用罗比达法则因为上下都是无穷小