Question

设f(u)连续,f(0)=0,f'(0)=1

设f(x)连续且f(0)=0,f'(0)=1 计算lim(x->0)=∫(t*f(x^2-t^2)dt)\x^4 积分的上下界x和0

Answer 1

∫(t*f(x^2-t^2)dt)= -0.5∫f(x^2-t^2)d(x^2-t^2)
设f(x)的一个原函数为 F(x),则上述积分等于 [F(x^2) - F(0)]/2
dF(x^2)/dx = 2xf(x^2)
而在 x=0处,dF(x^2)/dx = lim[F(x^2)-F(0)]/x,F(x^2)-F(0) = xdF(x^2)/dx
所以
原极限=lim[F(x^2) - F(0)]/2x^4 = lim dF(x^2)/dx /2x^3 = 2xf(x^2)/2x^3 = f(x^2)/x^2
而根据tailor一阶 展开 f(x^2)= f(0) + f'(0)x^2 = x^2
所以原极限 = x^2/x^2 =1