设f(x)在R上有定义, (0)=1, 且满足:-|||-lim _(xarrow 0)dfrac (ln (1-2x)+2xf(x))({x)^2}=0-|||-证明:f(x)在 x=0 处可导,并求f`(0).

f(x)在 [ -8,8] 上有定义, (0)=1, 且满足-|||-lim _(xarrow 0)dfrac (ln (1-2x)+2xf(x))({x)^

2.设f(x)在 x=0 的邻域内有定义, (0)=1, 且 lim _(xarrow 0)dfrac (ln (1+2x)-2xf(x))({x)^2}=0-

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f(x)可导,且满足 lim _(xarrow 0)dfrac (f(1)-f(1-2x))(x)=-2, 则曲线 =f(x) 在点(1,f(1))处的切线斜

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

3、设函数f(x)在x=0处连续，且lim_(xto0)(xf(x)-e^2sin x+1)/(ln(1+x)+ln(1-x))=-3证明f(x)在x=0处可导

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(

(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

[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