lim n bình + (-1) mũ n chia 2n bình + (-1 ) mũ n +1 04/11/2021 Bởi Audrey lim n bình + (-1) mũ n chia 2n bình + (-1 ) mũ n +1
$I=\lim\dfrac{n^2+(-1)^n}{2n^2+(-1)^{n+1}}$ * 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 $I=\dfrac{1}{2}$ Bình luận
$I=\lim\dfrac{n^2+(-1)^n}{2n^2+(-1)^{n+1}}$
* 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 $I=\dfrac{1}{2}$