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

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

设(x)=dfrac (1)(1-x), 求(x)=dfrac (1)(1-x)和(x)=dfrac (1)(1-x)设, 求和

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

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

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

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

1.求极限 lim _(xarrow 0)(dfrac (1)(ln (1+x))-dfrac (1)(x))

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

1.求极限 lim _(xarrow 0)(dfrac (1)(ln (1+x))-dfrac (1)(sin x))

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