$\lim u_n$
$=\lim\Big( \dfrac{\sin^42n}{n+2} +\dfrac{\cos^4n}{4n^2+8n}\Big)$
Ta có:
$0\le \dfrac{\sin^42n}{n+2}\le \dfrac{1}{n+2}$
$\lim 0=0$
$\lim\dfrac{1}{n+2}=\lim\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n}}=0$
Theo định lí kẹp, $\lim\dfrac{\sin^42n}{n+2}=0$
Tương tự, $\lim\dfrac{\cos^4n}{4n^2+8n}=0$
$\to \lim u_n=0$