已知 lim _(xarrow 0)([ 1+x+dfrac {f(x))(x)] }^dfrac (1{x)}=(e)^3, 则 lim _(xarrow 0
设lim _(xarrow 0)dfrac (ln (1+x+dfrac {f(x))(x))}(x)=3,则lim _(xarrow 0)dfrac (ln
设 lim _(xarrow 0)((1+x+dfrac {f(x))(x))}^dfrac (1{x)}=(e)^3 ,则 lim _(xarrow 0)((
若(0)=0, lim _(xarrow 0)dfrac (f(2x))(x)=4, 则 (0)=若
(B)若 lim _(xarrow 0)dfrac (f(x)+f(-x))(x) 存在,则 (0)=0.-|||-(C)若 lim _(xarrow 0)df
[题目]-|||-设 ((x)_(0))=3 则 lim _(xarrow 0)dfrac (f({x)_(0)+x)-f((x)_(0)-3x)}(x)= _
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
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)=(
极限lim _(xarrow 0)dfrac ({int )_(0)^(x^2)(e)^tdt}(2x)=()A、 lim _(xarrow 0)dfrac (
(3) lim _(xarrow 0)dfrac (1-cos 3x)({sin )^2x}