$I=\lim \Big(\dfrac{\sin n+\cos n}{n^2+1}\Big)$
$=\lim\Big( \dfrac{\sin n}{n^2+1}+\dfrac{\cos n}{n^2+1}\Big)$
Ta có:
$\Big| \dfrac{\sin n}{n^2+1}\Big| \le \dfrac{1}{n^2+1}$
$\to \lim\dfrac{\sin n}{n^2+1}=\lim\dfrac{1}{n^2+1}=\lim\dfrac{ \dfrac{1}{n^2}}{1+\dfrac{1}{n^2}}=0$
Tương tự, $\lim\dfrac{\cos n}{n^2+1}=0$
$\to I=0$