# How do you find the limit of 30 - (8/(6n^2)*(n+1)(2n+1)) - ((4/n)*(n+1)) as n approaches oo?

Mar 31, 2016

${L}_{n \to \infty} \left[3 - \left(\frac{8}{6 {n}^{2}} \left(n + 1\right) \left(2 n + 1\right)\right) - \left(\frac{4}{n} \left(n + 1\right)\right)\right] = - \frac{11}{3}$

#### Explanation:

${L}_{n \to \infty} \left[3 - \left(\frac{8}{6 {n}^{2}} \left(n + 1\right) \left(2 n + 1\right)\right) - \left(\frac{4}{n} \left(n + 1\right)\right)\right]$ can be written as

${L}_{n \to \infty} \left[3 - \left(\frac{8}{6} \left(1 + \frac{1}{n}\right) \left(2 + \frac{1}{n}\right)\right) - \left(4 \left(1 + \frac{1}{n}\right)\right)\right]$

We know that as $n \to \infty$, $\frac{1}{n} \to 0$

Hence ${L}_{n \to \infty} \left[3 - \left(\frac{8}{6} \left(1 + \frac{1}{n}\right) \left(2 + \frac{1}{n}\right)\right) - \left(4 \left(1 + \frac{1}{n}\right)\right)\right]$

= $\left[3 - \left(\frac{8}{6} \left(1 + 0\right) \left(2 + 0\right)\right) - \left(4 \left(1 + 0\right)\right)\right]$

= $\left(3 - \left(\frac{8}{6} \cdot 1 \cdot 2\right) - 4\right) = 3 - \frac{8}{3} - 4 = - \frac{11}{3}$