limx趋向于0,(1/sinx-1/x)=?

网友回答

用泰勒公式吧sinx在x=0展开
sinx=x-x^3/3!+x^5/5!-...(-1)^(k-1)*x^(2k-1)/(2k-1)!+...
则sinx~x
所以原式=lim(x-sinx)/xsinx
=lim[x^3/3!-x^5/5!+...(-1)^k*x^(2k-1)/(2k-1)!+...]/x*x
=lim[x/3!-x^3/5!+...(-1)^k*x^(2k-3)/(2k-1)!+...]
=0======以下答案可供参考======
供参考答案1:
这2个是等价无穷小，极限为0
供参考答案2:
lim(x->0)(1/sinx-1/x)
=lim(x->0)(x-sinx)/(xsinx) (0/0)
=lim(x->0)(1-cosx)/(xcosx+sinx) (0/0)
=lim(x->0)sinx/(-xsinx+2cosx)
=0供参考答案3:
lim (1/sinx - 1/x)
=lim (x-sinx) / (xsinx)
等价无穷小替换
=lim (x-sinx)/x^2
分子用泰勒展式
=lim (x-x+x^3/6+o(x^3))/x^2
=0