作业帮.证明当x→0时,[√(1+xsinx)-√(cosx)]～(3/4)x^2.当x→0时,[√(1+xsinx)-√(cosx)]～(3/4)x^2 lim[√(1+xsinx)-√(cosx)]/[(3/4)x^2] =lim(1+xsinx-cosx)/{[√(1+xsinx)+√(cosx)][(3/4)x^2]} =(2/3)lim(1+xsinx-cosx)/(x^2) =(2/3)lim(sinx+xcosx+s
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![.证明当x→0时,[√(1+xsinx)-√(cosx)]～(3/4)x^2.当x→0时,[√(1+xsinx)-√(cosx)]～(3/4)x^2 lim[√(1+xsinx)-√(cosx)]/[(3/4)x^2] =lim(1+xsinx-cosx)/{[√(1+xsinx)+√(cosx)][(3/4)x^2]} =(2/3)lim(1+xsinx-cosx)/(x^2) =(2/3)lim(sinx+xcosx+s](/uploads/image/z/12537388-28-8.jpg?t=.%E8%AF%81%E6%98%8E%E5%BD%93x%E2%86%920%E6%97%B6%2C%5B%E2%88%9A%281%2Bxsinx%29-%E2%88%9A%28cosx%29%5D%EF%BD%9E%283%2F4%29x%5E2.%E5%BD%93x%E2%86%920%E6%97%B6%2C%5B%E2%88%9A%281%2Bxsinx%29-%E2%88%9A%28cosx%29%5D%EF%BD%9E%283%2F4%29x%5E2+lim%5B%E2%88%9A%281%2Bxsinx%29-%E2%88%9A%28cosx%29%5D%2F%5B%283%2F4%29x%5E2%5D+%3Dlim%281%2Bxsinx-cosx%29%2F%7B%5B%E2%88%9A%281%2Bxsinx%29%2B%E2%88%9A%28cosx%29%5D%5B%283%2F4%29x%5E2%5D%7D+%3D%282%2F3%29lim%281%2Bxsinx-cosx%29%2F%28x%5E2%29+%3D%282%2F3%29lim%28sinx%2Bxcosx%2Bs)
.证明当x→0时,[√(1+xsinx)-√(cosx)]～(3/4)x^2.当x→0时,[√(1+xsinx)-√(cosx)]～(3/4)x^2 lim[√(1+xsinx)-√(cosx)]/[(3/4)x^2] =lim(1+xsinx-cosx)/{[√(1+xsinx)+√(cosx)][(3/4)x^2]} =(2/3)lim(1+xsinx-cosx)/(x^2) =(2/3)lim(sinx+xcosx+s
.证明当x→0时,[√(1+xsinx)-√(cosx)]～(3/4)x^2
.当x→0时,[√(1+xsinx)-√(cosx)]～(3/4)x^2
lim[√(1+xsinx)-√(cosx)]/[(3/4)x^2]
=lim(1+xsinx-cosx)/{[√(1+xsinx)+√(cosx)][(3/4)x^2]}
=(2/3)lim(1+xsinx-cosx)/(x^2)
=(2/3)lim(sinx+xcosx+sinx)/(2x)
=(1/3)lim[2(sinx)/x+cosx]=(1/3)(2+1)=1
所以当x→0时,[√(1+xsinx)-√(cosx)]～(3/4)x^2
我的问题是=(2/3)lim(1+xsinx-cosx)/(x^2) 到=(2/3)lim(sinx+xcosx+sinx)/(2x) 是怎么得来的
.证明当x→0时,[√(1+xsinx)-√(cosx)]～(3/4)x^2.当x→0时,[√(1+xsinx)-√(cosx)]～(3/4)x^2 lim[√(1+xsinx)-√(cosx)]/[(3/4)x^2] =lim(1+xsinx-cosx)/{[√(1+xsinx)+√(cosx)][(3/4)x^2]} =(2/3)lim(1+xsinx-cosx)/(x^2) =(2/3)lim(sinx+xcosx+s
其实就是分号上下同时对x求导
书上有定理的
你说你没学导数？这个问题根本不是高中问题！高中有等价无穷小吗？题目是大一的内容！最后一步楼上说得没错，就是分子分母同时求导，这是大学里的洛必达法则！
没学导数居然会问这么高深的问题！