樓上犯的是典型錯誤.
1、根據(jù)洛必達法則
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
這里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/2
2、
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