Hail Limits!

Using this formula for the sum of squares:

$\lim_{n\to\infty} \frac{n(1+4+9+16+\ldots+n^2)}{n^4+8n^3}$

$=\lim_{n\to\infty}\frac{n^2(n+1)(2n+1)}{6(n^4+8n^3)}$

$=\lim_{n\to\infty}\frac{2n^4+3n^3+n^2}{6n^4+48n^3}$

$=\lim_{n\to\infty}\frac{2+\frac3n+\frac1{n^2}}{6+\frac{48}n}$

$=\frac26\approx\boxed{0.33}.$

$\lim _{ n\rightarrow \infty }{ \frac { n\left( 1+4+9+16+...+{ n }^{ 2 } \right) }{ { n }^{ 4 }+8{ n }^{ 3 } } } =\lim _{ n\rightarrow \infty }{ \frac { \frac { 1 }{ { n }^{ 3 } } \sum _{ k=1 }^{ n }{ { k }^{ 2 } } }{ 1+\frac { 8 }{ n } } } \\ =\lim _{ n\rightarrow \infty }{ \frac { 1 }{ n } \sum _{ k=1 }^{ n }{ { \left( \frac { k }{ n } \right) }^{ 2 } } } =\int _{ 0 }^{ 1 }{ { x }^{ 2 } } dx=\frac { 1 }{ 3 } =0.333...\approx 0.33$