证明:若函数f(x)在x=0上连续,在(0,&)内可导,且当x趋向于0+时,lim f ' (x)=A.则f+'(x)存在且等于A.
人气:228 ℃ 时间:2019-08-18 00:53:26
解答
说明极限lim(x→0+) (f(x)-f(0))/x=A即可.由拉格朗日中值定理,f(x)-f(0)=f'(ξ)x,ξ介于0与x之间,且随着x在变.所以x→0+时,ξ→0+.
所以,lim(x→0+) (f(x)-f(0))/x=lim(x→0+) f'(ξ)=lim(ξ→0+) f'(ξ)=A,所以f+'(0)存在且等于A
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