Question

lim x→0 ln[1+f(x)/sinx]/a^x-1=A (a＞0，a≠1)，求lim x→0 f(x)/x^2=？

Answer 1

极限存在，分母为0分子也为0，故lim(1+f(x)/sinx)=1，limf(x)/sinx=0，f'(0)=0，f(x)比sinx高阶
limln(1+f(x)/sinx)/(a^x-1)
=lim(((f'(x)sinx-f(x)cosx)/sin²x)/(1+f(x)/sinx))/a^xlna
=lim(f'(x)sinx-f(x)cosx)/sin²x*lim1/(1+f(x)/sinx)a^xlna
=(1/lna)lim(f''(x)sinx+f'(x)cosx-f'(x)cosx+f(x)sinx)/2sinxcosx
=(1/lna)lim(f''(x)+f(x))/2cosx
=(1/lna)(f''(0)+f(0))/2
故f''(0)+f(0)=2Alna，又∵f(0)=0，∴f''(0)=2Alna
limf(x)/x^2=limf'(x)/2x=f''(0)/2=Alna

Answer 2

简单计算一下即可，答案如图所示