Question

lim n → ∞ (1^2+2^2+...+n^2)/4n^3+6n^2-5n+1

Answer 1

Step-by-step explanation:

We have,

$\lim _{n \rarr \infty } \frac{ {1}^{2} + {2}^{2} + {3}^{2} + ... + {n}^{2} }{4 {n}^{3} + 6 {n}^{2} - 5n + 1 }$

$= \lim _{n \rarr \infty } \frac{ \frac{n(n + 1)(2n + 1)}{6} }{4 {n}^{3} + 6 {n}^{2} - 5n + 1 }$

$= \frac{1}{6} \lim _{n \rarr \infty } \frac{( {n}^{2} + n)(2n + 1)}{4 {n}^{3} + 6 {n}^{2} - 5n + 1}$

$= \frac{1}{6} \lim _{n \rarr \infty } \frac{2 {n}^{3} + 3 {n}^{2} + n}{4 {n}^{3} + 6 {n}^{2} - 5n + 1 }$

$= \frac{1}{6} \lim _{n \rarr \infty } \frac{ {n}^{3} (2 + \frac{3}{n} + \frac{1}{ {n}^{2} } )}{ {n}^{3} (4 + \frac{6}{n} - \frac{5}{ {n}^{2} } + \frac{1}{ {n}^{3} }) }$

$= \frac{1}{6} \lim _{n \rarr \infty } \frac{ (2 + \frac{3}{n} + \frac{1}{ {n}^{2} } )}{ (4 + \frac{6}{n} - \frac{5}{ {n}^{2} } + \frac{1}{ {n}^{3} }) }$

$= \frac{1}{6} \times \frac{2 + 0 + 0}{4 + 0 - 0 + 0}$

$= \frac{1}{6} \times \frac{1}{2}$

$= \frac{1}{12}$