lim[ln(1-x)+sinxf(x)]/[e^(x^2)-1]=0,求f'(0)答案是1/2,求详解

Question

lim[ln(1-x)+sinxf(x)]/[e^(x^2)-1]=0,求f'(0)
答案是1/2,求详解

Answer 1

x趋于0的吧,首先在x趋于0时,e^x -1等价于x那么在这里分母e^(x^2)-1就等价于x^2所以原极限=lim(x趋于0) [ln(1-x)+sinx *f(x) ] /x^2 分子分母都趋于0,用洛必达法则同时求导=lim(x趋于0) [-1/(1-x)+cosx * f(x) +sinx ...