若(0)=0, lim _(xarrow 0)dfrac (f(2x))(x)=4, 则 '(0)=

(B)若 lim _(xarrow 0)dfrac (f(x)+f(-x))(x) 存在,则 (0)=0.-|||-(C)若 lim _(xarrow 0)df

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)=(

19.若 ((x)_(0))=-2 ,则 lim _(Delta xarrow 0)dfrac (f({x)_(0)+Delta x)-f((x)_(0))}(

14)/ lim _(xarrow 0)dfrac (x)(f(3x))=2, 则 lim _(xarrow 0)dfrac (f(2x))(x) 的值为 ()

极限lim _(xarrow 0)dfrac ({int )_(0)^(x^2)(e)^tdt}(2x)=()A、 lim _(xarrow 0)dfrac (

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

0°.求下列极限.-|||-(4) lim _(xarrow 0)(dfrac (1)({sin )^2x}-dfrac ({cos )^2x}({x)^2})

[题目]-|||-设 ((x)_(0))=3 则 lim _(xarrow 0)dfrac (f({x)_(0)+x)-f((x)_(0)-3x)}(x)= _

1.若f(x)在 x=0 处可导,且 (0)=0, 则 lim _(xarrow 0)dfrac (f(x))(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