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

7.证明:函数 =dfrac (1)(x)sin dfrac (1)(x) 在区间(0,1]内无界,但这函数不是x→0^+时的无穷大.

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

(int )_(0)^dfrac (pi {4)}dfrac (x)(1+cos 2x)dx=( ) .(int )_(0)^dfrac (pi {4)}dfr

已知曲线积分(int )_(1)^1(1+dfrac ({cos )^2}({x)^2}cos dfrac (y)(x))dx+(sin dfrac (y)(x

证明:ln dfrac (1+x)(1-x)+cos xgeqslant 1+dfrac ({x)^2}(2) -1lt xlt 1.证明:.

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

[例3.23]设-|||-in (0,1), 证明-|||-(1) (1+x)(ln )^2(1+x)lt (x)^2;-|||-(2) dfrac (1)(l

(5) lim _(xarrow 0)dfrac ({x)^2cos dfrac (1)(x)}(sin x) ,

(11) lim _(xarrow 0)dfrac (3sin x+{x)^2cos dfrac (1)(x)}((1+cos x)ln (1+x))= __

lim _(xarrow 0)dfrac ({(1+2{x)^2)}^dfrac (3{2)}-1}(tan 2x(cos sqrt {x)-1)}。。