求定积分极限

Question

设f(X)具有连续导数且f(0)=0,f’(0)=6求lim(x→0){∫上标为x^2下标为0f(t)dt/[{∫上标为x下标为0f(t)dt]^3}

Answer 1

lim(x→0)∫<0→x^2> f(t)dt/[∫<0→x> f(t)dt]^3
=lim(x→0)f(x^2)2x/[f(x)^3]（罗必塔法则）
=lim(x→0)f'(x^2)(2x)^2/[3f(x)^2*f'(x)]
=lim(x→0)f'(x^2)*4*[x/f(x)]^2/[3*f'(x)]
=lim(x→0)f'(x^2)*4/[3*f'(x)]*lim(x→0)[x/f(x)]^2
=f'(0)*4/[3*f'(0)]*[1/f'(0)]^2
=4/3*1/36
=1/27
方法是这样，不知是否计算正确