问题描述: 当x趋于0时lim(1/x2-1/tanx2) 的极限,咋做 1个回答 分类:数学 2014-10-09 问题解答: 我来补答 利用等价无穷小,洛必达法则求解.(x->0)lim(1/x^2-1/arctan^2(x))=(x->0)lim(1/x^2-cos^2(x)/sin^2(x))=(x->0)lim[sin^2(x)-x^2cos^2(x)]/[x^2sin^2(x)]=(x->0)lim(sinx+xcosx)(sinx-xcosx)/x^4=(x->0)lim[(sinx+xcosx)/x][(sinx-xcosx)/x^3]=(x->0)lim[(sinx)/x+cosx][(sinx-xcosx)]'/[x^3]'=(x->0)lim(1+1)[(cosx-cosx+xsinx)/(3x^2)]=(x->0)lim(2/3)(sinx)/x=2/3 展开全文阅读