Ta có $n\ge 1\to n+2>0$
$\lim\dfrac{ \sqrt{n^3+2n}+1}{n+2}$
$=\lim\dfrac{ \sqrt{n^2(n+\dfrac{2}{n})} +1}{n+2}$
$=\lim \dfrac{n\sqrt{n+\dfrac{2}{n}} +1}{n+2}$
$=\lim\dfrac{ \sqrt{n+\dfrac{2}{n}}+ \dfrac{1}{n} }{1+\dfrac{2}{n}}$
$=\lim \dfrac{ \sqrt{1+\dfrac{1}{n^2}} +\dfrac{1}{n\sqrt{n}} }{ \dfrac{1}{\sqrt{n}}+\dfrac{2}{n\sqrt{n}} }$
$=+\infty$
