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

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

计算lim _(xarrow 0)dfrac (ln (dfrac {1+x)(1-x))}((1+cos x)sin x)计算

设(x)=dfrac (1-x)(1+x), 则(x)=dfrac (1-x)(1+x)设,则

int dfrac (1)({x)^2}sqrt (dfrac {1-x)(1+x)}dx

23设 x<1, 且 neq 0 时,证明: dfrac (1)(x)+dfrac (1)(ln (1-x))lt 1.

设=ln sqrt (dfrac {1-x)(1-{x)^2}}则 dy|=ln sqrt (dfrac {1-x)(1-{x)^2}}设则dy|

、证明:当 -1lt xlt 0 时, arcsin sqrt (1-{x)^2}-arctan dfrac (x)(sqrt {1-{x)^2}}=dfrac

利用导数证明:当 gt 1 时, dfrac (ln (1+x))(ln x)gt dfrac (x)(1+x)

计算：lim _(xarrow 0)dfrac ({(1+x))^dfrac (2{x)}-(e)^2(1-ln (1+x))}(x)。计算：。

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