来自邵家玉的问题
lim[ln(1-x)+sinxf(x)]/[e^(x^2)-1]=0,求f'(0)答案是1/2,求详解
lim[ln(1-x)+sinxf(x)]/[e^(x^2)-1]=0,求f'(0)
答案是1/2,求详解
1回答
2020-08-0102:23
lim[ln(1-x)+sinxf(x)]/[e^(x^2)-1]=0,求f'(0)答案是1/2,求详解
lim[ln(1-x)+sinxf(x)]/[e^(x^2)-1]=0,求f'(0)
答案是1/2,求详解
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...