lim(x趋近于0) d((x-sinx)\xsinx) ÷ dx 怎么算?是求((x-sinx)\xsinx)的二阶导数后再求极限吗?
因
d[(x-sinx)/(xsinx)]/dx
= d(1/sinx - 1/x)/dx
= -cosx/(sinx)^2 + 1/x^2
= [(sinx)^2 -(x^2)cosx]/[(x^2)(sinx)^2],
故利用罗比达法则,有
g.e.= lim(x→0)[(sinx)^2 -(x^2)cosx]/[(x^2)(sinx)^2]
= lim(x→0)[(sinx)^2 -(x^2)cosx]/(x^4) (0/0)
= lim(x→0)[2sinxcosx + (x^2)sinx - 2xcosx]/(4x^3)
= (1/4)*lim(x→0)(sinx/x) + (1/2)lim(x→0)cosx*lim(x→0)[(sinx - x)/x^3]
= 1/4 + (1/2)*lim(x→0)[(sinx - x)/x^3] (0/0)
= 1/4 + (1/2)*lim(x→0)[(cosx - 1)/(3x^2)]
= 1/4 + (1/2)*(1/6)
= ……
再问: 你的 = lim(x→0)[(sinx)^2 -(x^2)cosx]/[(x^2)(sinx)^2]
= lim(x→0)[(sinx)^2 -(x^2)cosx]/(x^4)分母怎么变的?而且最终答案也不对啊!
再答: 这里用了等价无穷小替换 sinx ~ x (x→0); 最终答案可以再验算一下,方法应该没错。 g.e. = lim(x→0)[(sinx)^2 -(x^2)cosx]/[(x^2)(sinx)^2] = lim(x→0)[(sinx)^2 -(x^2)cosx]/(x^4) (0/0) = lim(x→0)[2sinxcosx + (x^2)sinx - 2xcosx]/(4x^3) = (1/4)*lim(x→0)(sinx/x) + (1/2)lim(x→0)cosx*lim(x→0)[(sinx - x)/x^3] = 1/4 + (1/2)*lim(x→0)[(sinx - x)/x^3] (0/0) = 1/4 + (1/2)*lim(x→0)[(cosx - 1)/(3x^2)] = 1/4 + (1/2)*(-1/6) = 1/6。