求lim x-0 sinx-x/x^2sinx的极限

lim {x→0} (sinx - x) / (x²sinx)
方法1：罗必塔
= lim {x→0} (cosx - 1) / (2xsinx + x²cosx) 上下求导
= lim {x→0} ( - sinx) / (2sinx + 2xcosx +2xcosx - x²sinx) 继续上下求导
= lim {x→0} ( - sinx) / (2sinx + 4xcosx - x²sinx) 上下同除sinx
= lim {x→0} ( - 1) / (2 + 4xcosx/sinx - x²) 利用sinx / x → 1
= - 1 / 6
方法2：等价无穷小替换 sinx = x - x³/6 + o(x³) -------就是取麦克劳林展开
lim {x→0} (sinx - x) / (x²sinx)
= lim {x→0} ( - x³/6) / (x²*x) = - 1 / 6

sinx-x/x^2sinx
洛必达上下同时求导(用三次)
=cosx-1/(2xsinx+x^2cosx)
=-sinx/(2sinx+2xcosx+2xcosx-x^2sinx)
=-cosx/(2cosx+4cosx-4xsinx-2xsinx-x^2cosx)
limx-0=-1/（2+4）=-1/6