lim (x→0) [(2x) / (1+x^2)]/sec x tan x+si
问题描述:
lim (x→0) [(2x) / (1+x^2)]/sec x tan x+si
lim (x→0) [(2x) / (1+x^2)]/sec x tan x+sin x怎么算出lim (x→0)(x/sin x ) (cos x^2/1+cos x^2)(2/1+x^2)
答
就化简一下就可以了lim (x→0) [(2x) / (1+x^2)] / (sec x tan x+sin x)=lim (x→0) [(2x) / (1+x^2)] / (sinx / cosx^2 +sin x)=lim (x→0) 2/(1+x^2) *x *cosx^2 /(sinx + sinx *cosx^2)=lim (x→0) x/sinx *cosx^...