Question

Знайти границі послідовності.

Answer 1

$\displaystyle \lim_{n\to \infty}\left(\frac{n^3+1}{n+1}-\frac{n^3-3n}{n^2}\right)=\\\\\lim_{n\to \infty}\left(\frac{n^3+1}{n+1}-\frac{n(n^2-3)}{n^2}\right)=\\\\\lim_{n\to \infty}\left(\frac{n^3+1}{n+1}-\frac{n^2-3}{n}\right)=$

$\displaystyle \lim_{n\to \infty}\left(\frac{n(n^3+1)}{n(n+1)}-\frac{(n+1)(n^2-3)}{n(n+1)}\right)=\\\\\lim_{n\to \infty}\frac{n(n^3+1)-(n+1)(n^2-3)}{n(n+1)}=\\\\\lim_{n\to \infty}\frac{n^4+n-(n^3-3n+n^2-3)}{n(n+1)}=\\\\\lim_{n\to \infty}\frac{n^4+n-n^3+3n-n^2+3}{n(n+1)}=\\\\\lim_{n\to \infty}\frac{n^4-n^3-n^2+4n+3}{n(n+1)}=\\\\\lim_{n\to \infty}\frac{n^4-n^3-n^2+n+3n+3}{n(n+1)}=$

$\displaystyle \lim_{n\to \infty}\frac{n^3(n-1)-n(n-1)+3n+3}{n(n+1)}=\\\\\lim_{n\to \infty}\frac{(n-1)(n^3-n)+3(n+1)}{n(n+1)}=\\\\\lim_{n\to \infty}\frac{n(n-1)(n^2-1)+3(n+1)}{n(n+1)}=\\\\\lim_{n\to \infty}\frac{n(n-1)(n-1)(n+1)+3(n+1)}{n(n+1)}=\\\\\lim_{n\to \infty}\frac{(n+1)(n(n-1)(n-1)+3)}{n(n+1)}=\\\\\lim_{n\to \infty}\frac{n(n-1)(n-1)+3}{n}=$

$\displaystyle \lim_{n\to \infty}\frac{n(n^2-2n+1)+3}{n}=\\\\\lim_{n\to \infty}\frac{n^3-2n^2+n+3}{n}=\\\\\lim_{n\to \infty}\frac{n^3(\frac{n^3}{n^3}-\frac{2n^2}{n^3}+\frac{n}{n^3}+\frac{3}{n^3})}{n}=\\\\\lim_{n\to \infty} \left(n^2(1-\frac{2}{n}+\frac{1}{n^2}+\frac{3}{n^3})\right)=+\infty$