求极限suppose that f(x) is differential,and f(0)=0,F(x)=∫<0,x>f[(x^n)-(t^n)]dt,compute limF(x)/(x^2n)
1。n=1,
F(x)=∫f(x-t)dt=(u=x-t)
=∫f(u)du
==>
limF(x)/(x^2) =
=limF'(x)/(x^2)'=
=limf(x)/(2x)=
=(1/2)lim[f(x)-f(0)]/(x)=
=f'(0)/2。
2。n>1,好象要加条件。
如:设f”(x) 存在,且f”(x)有界。
F(x)=∫f[(x^n)-(t^n)]dt=(分部积分)
=f[(x^n)-(x^n)]x-f[(x^n)-(0^n)]0-
-∫t[(x^n)-(t^n)]’f’[(x^n)-(t^n)]dt=
=n∫(t^n)f’[(x^n)-(t^n)]dt=
=n/(n...全部
1。n=1,
F(x)=∫f(x-t)dt=(u=x-t)
=∫f(u)du
==>
limF(x)/(x^2) =
=limF'(x)/(x^2)'=
=limf(x)/(2x)=
=(1/2)lim[f(x)-f(0)]/(x)=
=f'(0)/2。
2。n>1,好象要加条件。
如:设f”(x) 存在,且f”(x)有界。
F(x)=∫f[(x^n)-(t^n)]dt=(分部积分)
=f[(x^n)-(x^n)]x-f[(x^n)-(0^n)]0-
-∫t[(x^n)-(t^n)]’f’[(x^n)-(t^n)]dt=
=n∫(t^n)f’[(x^n)-(t^n)]dt=
=n/(n+1)∫f’[(x^n)-(t^n)]d(t^(n+1))=
=[n/(n+1)]f’[0]x^(n+1)+
+n^2/(n+1)∫t^(2n)f”[(x^n)-(t^n)]dt
由于
|∫t^(2n)f”[(x^n)-(t^n)]dt|≤
≤M|∫t^(2n)]dt=M|x|^(2n+1)
==》
lim∫t^(2n)f”[(x^n)-(t^n)]dt/(x^2n) =0
==》
limF(x)/(x^2n) =
=lim[n/(n+1)]f’[0]x^(n+1)(x^2n) =
=0,当f’[0]=0
=∞,当f’[0]≠0。
这题可能应为求limF(x)/(x^(n+1))=
=[n/(n+1)]f’[0]。
。收起