设 函数 f ( x ) 在 x = 1 处可导且lim _(xarrow 0)dfrac (f(1)-f(1-x))(2x)=1则 lim _(xarrow 0)dfrac (f(1)-f(1-x))(2x)=1

1.设f(x)为可导函数,且 lim _(xarrow 0)dfrac (f(1)-f(1-x))(2x)=-1 ,则 f(1)= __

设 函数 f ( x ) 在 x = 0 处可导，且lim _(xarrow 0)dfrac (f(2x)-f(0))(ln (1+3x))=1，则f(0)=(

没f`(x)存在,且 lim _(xarrow 0)dfrac (f(1)-f(1-x))(3x)=1, 则 (1)=..

设 lim _(xarrow 0)((1+x+dfrac {f(x))(x))}^dfrac (1{x)}=(e)^3 ,则 lim _(xarrow 0)((

(2)设函数f(x)在区间 (-1,1) 内有定义,且 lim _(xarrow 0)f(x)=0, 则-|||-(A)当 lim _(xarrow 0)dfr

已知f(x)满足 lim _(xarrow 1)dfrac (f(x))(ln x)=1, 则 () .-|||-(A) f(1)=0 (B) lim _(xa

已知 lim _(xarrow 0)([ 1+x+dfrac {f(x))(x)] }^dfrac (1{x)}=(e)^3, 则 lim _(xarrow 0

1．设f(x)在x=0处可导，且f(0)=0,则lim _(xarrow 0)dfrac (f(3x)-f(x))(x)-|||-__=______.1．设f(

设lim _(xarrow 0)dfrac (ln (1+x+dfrac {f(x))(x))}(x)=3，则lim _(xarrow 0)dfrac (ln

2.设函数f(x)在区间 (-1,1) 内有定义,且 lim _(xarrow 0)f(x)=0, 则 ()-|||-A.当 lim _(xarrow 0)df