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)=(
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
已知 lim _(xarrow 0)([ 1+x+dfrac {f(x))(x)] }^dfrac (1{x)}=(e)^3, 则 lim _(xarrow 0
设 (x+dfrac (1)(x))=(x)^2+dfrac (1)({x)^2}, 则 lim _(xarrow 3)f(x)= __
设lim _(xarrow 0)dfrac (ln (1+x+dfrac {f(x))(x))}(x)=3,则lim _(xarrow 0)dfrac (ln
[题目]已知f((x)在 x=0 处可导,且 (0)=0, 则-|||-lim _(xarrow 0)dfrac ({x)^2f(x)-2f((x)^3)}({
1.设f(x)为可导函数,且 lim _(xarrow 0)dfrac (f(1)-f(1-x))(2x)=-1 ,则 f(1)= __
1.-|||-若limf(x)存在,且 (x)=(x)^3+dfrac (2{x)^2+1}(x+1)+2lim _(xarrow 1)f(x) ,则 lim
[题目]已知f(x)在 x=0 处可导,且 (0)=0, 则-|||-lim _(xarrow 0)dfrac ({x)^2f(x)-2f((x)^3)}({x
(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