Question

设f(x)可导,且f(0)=0,F(x)=∫﹙0→x﹚{[t^(n-1)]f(x^n-t^n)}dt,求lim (x→0)[f(x)/x^2n](n为自然数)

Answer 1

结果是f'(0)/(2n)
题目应该是计算lim{x→0}F(x)/x^(2n)吧？
F(x)=1/n∫{0,x}f(x^n-t^n)d(t^n)
=1/n∫{x^n,0}f(z)*(-1)d(z) 令x^n-t^n=z,则t^n=x^n-z
=1/n∫{0,x^n}f(z)d(z)
lim{x→0}F(x)/x^(2n)=lim{x→0}[1/n*f(x^n)*n*x^(n-1)]/[2*n*x^(2n-1)]
=lim{x→0}f(x^n)/(2*n*x^n)
=lim{x→0}[f(x^n)-f(0)/[2*n*(x^n-0)]
=1/(2n)*lim{x→0}[f(x^n)-f(0)/(x^n-0)
=1/(2n)*f'(0)