设 lim _(xarrow 0)((1+x+dfrac {f(x))(x))}^dfrac (1{x)}=(e)^3 ,则 lim _(xarrow 0)((1+dfrac {f(x))(x))}^dfrac (1{x)}= __

设lim _(xarrow 0)dfrac (ln (1+x+dfrac {f(x))(x))}(x)=3，则lim _(xarrow 0)dfrac (ln

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

[例15]设f(x)连续可导,且 lim _(xarrow 0)([ 1+x+dfrac {f(x))(x)] }^dfrac (1{x)}=(e)^3, 求f

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f(x)满足 lim _(xarrow 1)dfrac (f(x))(ln x)=1, 则 () .-|||-(A) f(1)=0 (B) lim _(xa

设 (x+dfrac (1)(x))=(x)^2+dfrac (1)({x)^2}, 则 lim _(xarrow 3)f(x)= __

(9)已知 lim _(xarrow a)f(x)=lim _(xarrow a)g(x), 则 lim _(xarrow a)dfrac (f(x))(g(x

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

(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

[例8]设f(x )在 x=1 连续,且 lim _(xarrow 1)dfrac (f(x)-2)(x-1) 存在,则 (1)= __-|||-解:因为 li