当x趋向于0时,x-sinx所表示的无穷小量是x的几阶无穷小?
人气:135 ℃ 时间:2020-02-04 23:54:02
解答
如果知道L'Hospital法则就好办
lim{x->0} (x-sinx)/x^k
=lim{x->0} (1-cosx)/(k*x^(k-1))
=lim{x->0} sinx/(k(k-1)*x^(k-2))
当且仅当k=3时极限存在且非0.
如果知道Taylor公式也好办
sinx=x-x^3/6+o(x^4)
显然有x-sinx是3阶无穷小.
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