求下列数列的极限,

lim(n->inf)[3n^2+n]/[2n^2-1] = lim(n->inf)[3+1/n]/[2-1/n^2] = 3/2
【当分子,分母都是无穷大时.分子,分母同除以一个无穷大因子.使得分子,分母中至少有1个不再是无穷大.极限就出来了.】
lim(n->inf)[(3n)^2+n]/[(2n)^2-1] = lim(n->inf)[9+1/n]/[4-1/n^2] = 9/4
lim(n->inf)[1+1/2+...+1/2^n]/[1+1/4+...+1/4^n]
=lim(n->inf)[1-1/2^(n+1)]/(1-1/2)*(1-1/4)/[1-1/4^(n+1)]
=lim(n->inf)[1-1/2^(n+1)]/[1-1/4^(n+1)]*(4-1)/(4-2)
= 3/2
【分子,分母是无限项和时.先分别求有限项和,再算极限】