证明:一定存在 _(0)in [ 0,dfrac (1)(2)] , 使得 ((x)_(0))=-|||-((x)_(0)+dfrac (1)(2)).

设函数f(x)在[0,1]二阶可导，(0)=2 (1)=1, (dfrac (1)(2))=0，证明至少存在一点(0)=2 (1)=1, (dfrac (1)(

39.设f(x)在[0,1]上连续,在(0,1)内可导,且 f(0)=0 ,(dfrac (1)(2))=2, (1)=dfrac (1)(2),-|||-(1

2、设f(x)在区间[0,1]上可导, (1)=2(int )_(0)^dfrac (1{2)}(x)^2f(x)dx, 证明:存在 varepsilon in

4.设函数f(x)在(1)=dfrac (1)(2),11上连续，（0，1）内可导，f(0)=0(1)=dfrac (1)(2),11,证明存在一点(1)=df

(2)证明: (x)=dfrac (1)(x)cos dfrac (1)(x) 在(0,1)内无界。

lim _(xarrow 0)dfrac ({x)^2sin dfrac (1)(x)}(tan x)=A.0B.1C.lim _(xarrow 0)dfrac

(sin x)=dfrac (1)({cos )^2x} in (0,dfrac (pi )(2)),则(sin x)=dfrac (1)({cos )^2x}

(int )_(0)^1arctan dfrac (x)(2)dx=()(int )_(0)^1arctan dfrac (x)(2)dx=()(int )_(

当x→0时,变量 dfrac (1)({x)^2}sin dfrac (1)(x) 是-|||-__当x→0时,变量 dfrac (1)({x)^2}sin d

1.证明下列各式:-|||-(1) -(x)^2=0(x)(xarrow 0);-|||-(2) sin sqrt (x)=O((x)^dfrac (3{2)}