Question

помогииииииииииите)))

Answer 1

так как 1/n при n->+oo стр к 0 то 
$frac{2n+1}{n-1}=frac{2+frac{1}{n}}{1-frac{1}{n}}=2$

$frac{n^2}{2n^2-1}=frac{1}{2-frac{1}{n^2}}=frac{1}{2}$

Answer 2

$lim_{n to +infty} cfrac{2n+1}{n-1} =lim_{n to +infty} cfrac{ frac{2n}{n} +frac{1}{n}}{ frac{n}{n} -frac{1}{n}}= lim_{n to +infty} cfrac{2+frac{1}{n}}{1-frac{1}{n}}= cfrac{2}{1} =2 \\ lim_{n to +infty} cfrac{n^2}{2n^2-1}= lim_{n to +infty} cfrac{ frac{n^2}{n^2} }{ frac{2n^2}{n^2} -frac{1}{n^2}}= lim_{n to +infty} cfrac{ 1 }{2 -frac{1}{n^2}}=cfrac{1}{2}=0.5$