设y=f(x) 在x0处可导,且 '((x)_(0))=2, 则lim _(xarrow 0)dfrac (f({x)_(0)+2)x-f((x)_(0)-f(x)}(Delta x)= __-|||-A 6-|||-B) -6-|||-C) dfrac (1)(6)-|||-D) -dfrac (1)(6)-|||-__

(6)设f(x)在x0处可导,则 lim _(harrow 0)dfrac (f({x)_(0)+h)-f((x)_(0)-h)}(h)= () ;-|||-(

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 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)=(

154 设 lim _(xarrow {x)_(0)^+}f(x)=lim _(xarrow {x)_(0)^-}(x)=a, 则-|||-(A)f(x)在 =

[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

[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)}({

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

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

[题目]设f(x)在 x=0 处连续,且 lim _(xarrow 0)dfrac ((f(x)+1){x)^2}(x-sin x)=2,-|||-则曲线 =f