Mn cho e hỏi sao cách trên sai ạ
\(\lim\limits_{x\rightarrow+\infty}\dfrac{1^2+2^2+3^2+....+n^2}{n^3}=\lim\limits_{x\rightarrow+\infty}=\left(\dfrac{1^2}{n^3}+\dfrac{2^2}{n^3}+...+\dfrac{n^2}{n^3}\right)=0\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{1^2+2^2+3^2+...+n^2}{n^3}=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}}{n^3}=\dfrac{1}{3}\)