3.若函数 f(x)= {(1-dfrac {x)(2))}^dfrac (1{x)} xlt 0 x+a xgeqslant 0

14.设 f(x)= ^x) xlt 0 dfrac (1)(1+x) xgeqslant 0f(x-1)dx

若f(x)= ) 1-x,xgeqslant 0 (2)^x,xlt 0 .________________。若，则________________。

A X=0 =1-dfrac (theta )(100), X=1 =dfrac (theta )(100). B X=0 =1-dfrac (theta

设函数 f(x)= ^2)(int )_(0)^xtan xdt xlt 0 1 x=0 dfrac (1)({x)^2}(int )_(0)^xsin

(2) lim _(xarrow 0)dfrac ({(1-dfrac {1)(2)(x)^2)}^dfrac (2{3)}-1}(xln (1+x))

函数 f(x)= ) 3x-1 xgeqslant 1 (x)^2 xlt 1f(x)=()

求函数f(x)= ) 2(x+1),xlt 0 (e)^2x+1,xgeqslant 0 .的反函数.求函数的反函数.

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

3、已知 lim _(xarrow 0)((1-dfrac {x)(2))}^dfrac (a{x)}=(e)^2, 则常数 a= __

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