$I=\lim\dfrac{n^2+(-1)^n}{2n^2+(-1)^{n+1}}$
* Khi $n+1$ lẻ $\Leftrightarrow$ $n$ chẵn:
$\lim\dfrac{n^2+1}{2n^2+2}$
$=\lim\dfrac{1+\dfrac{1}{n^2}}{2+\dfrac{2}{n^2}}$
$=\dfrac{1}{2}$
* Khi $n+1$ chẵn $\Leftrightarrow$ $n$ lẻ:
$\lim\dfrac{n^2-1}{2n^2}$
$=\lim\dfrac{1-\dfrac{1}{n^2}}{2}$
$=\dfrac{1}{2}$
Vậy $I=\dfrac{1}{2}$