怎么得来的？求详细过程

当n→+∞时，ln(1+3ⁿ)→+∞，[ln(1+3ⁿ)]/n属于0/0型，用罗比塔法则变换。n'=1ln'(1+3ⁿ)=(1+3ⁿ)'/(1+3ⁿ)=(ln3)3ⁿ/(1+3ⁿ)lim【n→+∞】e^{[ln(1+3ⁿ)]/n}=e^lim【n→+∞】{[ln(1+3ⁿ)]/n}=e^lim【n→+∞】[(ln3)3ⁿ/(1+3ⁿ)]=e^(ln3)lim【n→+∞】[3ⁿ/(1+3ⁿ)]=3