原式为:lim _(xarrow 0)(dfrac (1)({e)^x-1}-dfrac (1)(ln (1+x)))= )=原式为:
lim _(xarrow 1)(dfrac (x)(x-1)-dfrac (1)(ln x))-|||-__ __
lim _(xarrow 1)(dfrac (x)(x-1)-dfrac (1)(ln x))-|||-__ __
lim _(xarrow 0)(dfrac (1)(x)-dfrac (1)({e)^x-1})-|||-__;;
lim _(xarrow 0)(dfrac (1)(x)-dfrac (1)({e)^x-1}) ()-|||-__ __
lim _(xarrow 0)(dfrac (1)(x)-dfrac (1)({e)^x-1}) ()-|||-__ __
[题目] lim _(xarrow 1)(dfrac (x)(x-1)-dfrac (1)(ln x))
lim _(xarrow 1)(e)^dfrac (1{x-1)}=
lim _(xarrow 1)dfrac ({x)^2-1}(x-1)(e)^dfrac (1{x-1)}=
解:lim _(xarrow 1)(dfrac (1)(x-1)-dfrac (1)(ln x))-|||-__ __解: