Answer
$\frac{1}{3}$
Work Step by Step
Find the limit $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}(\frac{i}{n})^{2}$
$\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}(\frac{i}{n})^{2}=\lim\limits_{n \to \infty}\frac{1}{n^{3}}\sum\limits_{i =1}^{n}({i}^{2})$
Since, $ \sum \limits_{i =1}^{n}i^{2}=[\frac{n(n+1)(2n+1)}{6}]$
Thus,
$\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}(\frac{i}{n})^{2}=\lim\limits_{n \to \infty}[\frac{1}{n^{3}}(\frac{n(n+1)(2n+1)}{6})]$
$=\lim\limits_{n \to \infty}[\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^{2}}]$
$=\frac{1}{3}+0+0$
Hence, $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}(\frac{i}{n})^{2}=\frac{1}{3}$