0 (1/x^2-1/(arctanx)^2) 1、e/2 2、 -2/3

0 (1/x^2-1/(arctanx)^2) 1、e/2 2、 -2/3

楼上犯的是典型错误.1、根据洛必达法则lim( x->0) [(1+x)^1/x-(1+2x)^1/2x]/sinx =lim (x->0){ [(1+x)^1/x]g(x)-[(1+2x)^1/2x]h(x)}/cosx这里g(x)=-(1/x^2)ln(1+x)+1/[x(1+x)] h(x)=(1/2){[-(1/x^2)ln(1+2x)]+2/[x(1+2x)]} lim( x->0)g(x)= lim( x->0){[-(1/x^2)ln(1+x)]+1/[x(1+x)]}=lim( x->0)[-ln(1+x)/2x]=-1/2 lim( x->0)h(x)= lim( x->0)(1/2){[-(1/x^2)ln(1+2x)]+2/[x(1+2x)]} =lim( x->0)(1/2)[-2ln(1+2x)/2x]=-1∴原式=e(-1/2)-e(-1)=e/22、lim( x->0) (1/x^2-1/(arctanx)^2) =lim(x->0)(arctanx²-x²)/(x²arctanx²)=lim(x->0)(arctanx²-x²)/(x^4)=lim(x->0)[(2arctanx)/(1+x^2)-2x]/4x^3=lim(x->0)[(arctanx-x-x^3]/2x^3=lim(x->0){[1/(1+x^2)]-1-3x^2}/6x^2=lim(x->0)(-4x^2-3x^4)/6x^2=-2/3======以下答案可供参考======供参考答案1:1、lim x->0 [(1+x)^1/x-(1+2x)^1/2x]/sinx =lim x->0(1+x-1-2x)/x=-12、lim x->0 (1/x^2-1/(arctanx)^2) =limx->0(arctanx²-x²)/(x²arctanx²)=limx->0(1-1)/x²=0

这个问题的回答的对