Question

初学微积分，求助啊求助！！！ x趋向于0时，求lim(1/ln(1+x) +1/ln(1-x) )

x趋向于0时，求lim(1/ln(1+x) +1/ln(1-x) )
x趋向于0时，求lim(3sinx+x^2cos(1/x))/((1+cosx)ln(1+x))

Answer 1

x趋向于0时，lim(1/ln(1+x) +1/ln(1-x) )=lim[(1/ln(1+x))*(1/ln(1-x))]=lim{[ln(1+x)+ln(1-x)]/[ln(1+x)*ln(1-x)]
=lim{ln(1-x^2)/[ln(1+x)*ln(1-x)]}
=lim([-2x/(1-x^2)]/{[(ln(1-x))/(1+x)]-[(ln(1+x))/(1-x)]})
=-2lim{x/ln[(1-x)/(1+x)]}
=-2/lim{1/[(1+x)/(1-x)]}
=-2；

x趋向于0时，lim(3sinx+x^2cos(1/x))/((1+cosx)ln(1+x))
=lim{[3cosx+2xcos(1/x)-sin(1/x)]/[(1+cosx)/(1+x)-sinxln(1+x)]}
=lim{[3+2xcos(1/x)-sin(1/x)]/[2-sinxln(1+x)]}
=3/2+lim{2xcos(1/x)-sin(1/x)}/2
=3/2+lim{2xcos(1/x)}-lim{sin(1/x)}/2
=3/2+0-lim{sin(1/x)}/2
=3/2-lim{sin(1/x)}/2
由于当x趋向于0时，sin(1/x)取值在-1~1之间不固定，故原式极限不存在。