$u_n=-2n^2-3n+4$
$=-2\Big( n^2+\dfrac{3}{2}n-2\Big)$
$=-2\Big(n^2+2.n.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{41}{16}\Big)$
$=-2\Big(n+\dfrac{9}{16}\Big)^2+\dfrac{41}{8}\le \dfrac{41}{8}$
$\to (u_n)$ bị chặn trên.
$\lim u_n=\lim n^2\Big(-2-\dfrac{3}{n}+\dfrac{4}{n^2}\Big)=-\infty$
$\to(u_n)$ không bị chặn dưới.