设f(x)是可导函数且f(0)=0,F(x)=∫t^(n-1)f(x^n-t^n)dt,求lim(x→+∞)F(x)/x^2n设f(x)是可导函数,且f(0)=0,F(x)=∫t^(n-1)f(x^n-t^n)dt,求lim(x→+∞)F(x)/x^2n答案是f'(0)/2n求详解

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设f(x)是可导函数且f(0)=0,F(x)=∫t^(n-1)f(x^n-t^n)dt,求lim(x→+∞)F(x)/x^2n设f(x)是可导函数,且f(0)=0,F(x)=∫t^(n-1)f(x^n-t^n)dt,求lim(x→+∞)F(x)/x^2n答案是f'(0)/2n求详解
设f(x)是可导函数且f(0)=0,F(x)=∫t^(n-1)f(x^n-t^n)dt,求lim(x→+∞)F(x)/x^2n
设f(x)是可导函数,且f(0)=0,F(x)=∫t^(n-1)f(x^n-t^n)dt,求lim(x→+∞)F(x)/x^2n
答案是f'(0)/2n
求详解

设f(x)是可导函数且f(0)=0,F(x)=∫t^(n-1)f(x^n-t^n)dt,求lim(x→+∞)F(x)/x^2n设f(x)是可导函数,且f(0)=0,F(x)=∫t^(n-1)f(x^n-t^n)dt,求lim(x→+∞)F(x)/x^2n答案是f'(0)/2n求详解
你写的式子我看不太明白,但是我觉得肯定是用洛必达法则做的,你试一试!

用泰勒公式

题目应该是极限x趋向0。先用分部积分得到F(x)=∫[0-x]t^(2n-1)*f‘(x^n-t^n)dt,再换元后用L'Hospital法则
换元:u^n=x^n-t^n
F(x)=∫[0-x](x^n-u^n)*u^(n-1)*f'(u^n)du
连续用两次L'Hospital法则即可