求ln(1+ x)的无穷近似值．

Question

如题所述

Answer 1

无穷近似值limx/ln(x+1)=1，可推x~ln(1+x)，xlna~ln(1+xlna)，a^x~l+xlna
=lime^(ln(2-3^((arctan根号x)^2))2/sinx))
=lime^(1-3^((arctan根号x)^2))2/sinx)
=lime^(1-(1+(arcsin(根号x/(x+1)))^2ln3))2/sinx)
=lime^(-2(根号x/(x+1))^2ln3)/sinx)
=lime^(-2ln3*(x/sinx)/(x+1)^2)
=e^(-2ln3)
=1/3e^2