Limit

$\begin{aligned} L & = \lim_{n\to\infty}\frac{1^2+2^2+3^2+\cdots +n^2}{n^3} \\ & = \lim_{n\to\infty}\displaystyle\sum_{k=1}^{n}\frac{k^2}{n^3} \\& = \lim_{n\to\infty}\frac{1}{n}\displaystyle\sum_{k=1}^{n}\frac{k^2}{n^2} \\& = \displaystyle\int_{0}^{1} x^2\,dx \phantom{ccccccc} {\color{#3D99F6}\text{By Riemann Sums}}\\& = \frac{1}{3} \\& =\boxed{0.333}\end{aligned}$

$SOLUTION$

$\large\displaystyle\lim_{n->\infty} \frac{1^2 + 2^2 + 3^2 + \cdots+ n^2}{n^3}$

$\implies$ $\displaystyle\lim_{n->\infty} \frac{n(n+1)(2n+1)}{6n^3}$

$\implies$ $\displaystyle\lim_{n->\infty} \frac{n^3(1+\frac{1}{n})(2+\frac{1}{n})}{6n^3}$

$\implies$ $\displaystyle\lim_{n->\infty} \frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}$

$\implies$ $\displaystyle \frac{(1) \times (2)}{6}$

$\implies$ $\displaystyle \frac{2}{6}$

$\implies$ $\displaystyle \frac{1}{3} = \boxed{0.333}$

Similar solution with @Naren Bhandari's

$\begin{aligned} L & = \lim_{n \to \infty} \frac {1^2+2^2+3^2+\cdots + n^2}{n^3} \\ & = \lim_{n \to \infty} \frac 1n \sum_{k=1}^n \left(\frac kn\right)^2 & \small \color{#3D99F6} \text{Using Riemann sums: } \\ & = \int_0^1 x^2 \ dx \\ & = \frac {x^3}3 \ \bigg|_0^1 = \frac 13 \approx \boxed{0.333} \end{aligned}$

$\large\displaystyle=\lim_{n \to \infty} \frac{\sum_{k=1}^{n}k^2}{n^3}$ $\large\displaystyle=\lim_{n \to \infty} \frac{\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}}{n^3}$ $\large\displaystyle=\lim_{n \to \infty} \frac{\frac{n^2}{3}+\frac{n}{2}+\frac{1}{6}}{n^2}$ $\large\displaystyle=\lim_{n \to \infty}\left(\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^2}\right)$ $\large\displaystyle=\frac{1}{3}+\lim_{n \to \infty}\left(\frac{1}{2n}+\frac{1}{6n^2}\right)$ Because $\large\displaystyle\lim_{n \to \infty}2n=\infty$ and $\large\displaystyle\lim_{n \to \infty}6n^2=\infty$ , the new limit becomes $\large\displaystyle\lim_{n \to \infty}\left(\frac{1}{2n}+\frac{1}{6n^2}\right)=\lim_{n \to \infty}\left(\frac{1}{\infty}+\frac{1}{\infty}\right)=0$ . With this, we arrive at our solution: $\large\displaystyle=\frac{1}{3}=0.333\dots$

Relevant wiki: Stolz–Cesàro theorem

$\begin{aligned} L&=\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + \cdots+ n^2}{n^3} &\small \color{#3D99F6}\text{By Stolz-Cesàro theorem} \\&=\lim_{n\to\infty} \frac{(n+1)^2}{(n+1)^3-n^3} \\&=\lim_{n\to\infty} \frac{n^2+2n+1}{3n^2+3n+1} \\&=\lim_{n\to\infty} \frac{1+\frac 2n+\frac 1{n^2}}{3+\frac 3n+\frac 1{n^2}} \\&=\frac 13\approx\boxed{0.333} \end{aligned}$