12.这对你们来说不是难题,可对我来说是很头疼的问题,

limn[ln(n+1)-lnn]
=limn*ln[(n+1)/n]
=limln[(n+1)/n]^n
=limln[1+(1/n)]^n
=lne
=1
lim[(x+2a)/(x-a)]^(x/3)=8
===> lim[(x-a+3a)/(x-a)]^(x/3)=8
===> lim[1+3a/(x-a)]^(x/3)=8
===> lim{[1+3a/(x-a)]^(x-a)/(3a)}^[ax/(x-a)]=8
===> e^lim[ax/(x-a)]=8
===> e^a=8
===> a=ln8=3ln2