Can't Figure Out Missing Step: lim[n->infty] (3/n^3) [n(n+1)(2n+1)]/[6] + (3/n)n

pipsydoodles said:

The example is as follows:

. . .\(\displaystyle \displaystyle \lim_{n \rightarrow \infty} \dfrac{3}{n^3}\, \dfrac{n\, (n\, +\, 1)\, (2n\, +\, 1)}{6}\, +\, \dfrac{3}{n}\, n\)

. . .\(\displaystyle \displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{2}\, \cdot\, \dfrac{n}{n}\, \cdot\, \left(\dfrac{n\, +\, 1}{n}\right)\, \left(\dfrac{2n\, +\, 1}{n}\right)\, +\, 3\)

. . .\(\displaystyle \displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{2}\, \cdot\, 1\, \left(1\, +\, \dfrac{1}{n}\right)\, \left(2\, +\, \dfrac{1}{n}\right)\, +\, 3\)

. . . . . . . . . .\(\displaystyle \displaystyle =\, \dfrac{1}{2}\, \cdot\, 1\, \cdot\, 1\, \cdot\, 2\, +\, 3\, =\, 4\)

Where I start to get confused is on the second line as it appears that some intermediary steps were skipped for brevity. Normally, I would have solved the question like so:

. . .\(\displaystyle \displaystyle \lim_{n \rightarrow \infty} \dfrac{3}{n^3}\, \dfrac{n\, (n\, +\, 1)\, (2n\, +\, 1)}{6}\, +\, \dfrac{3}{n}\, n\)

. . .\(\displaystyle \displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{2}\, \cdot\, \dfrac{1}{n^3}\, \cdot\, \left(2n^3\, +\, 3n^2\, +\, n\right)\, +\, 3\)

. . .\(\displaystyle \displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{2}\, \cdot\, \left(2\, +\, \dfrac{3}{n}\, +\, \dfrac{1}{n^2}\right)\, +\, 3\)

. . . . . . . . . .\(\displaystyle \displaystyle =\, \dfrac{1}{2}\, \cdot\, 2\, +\, 3\, =\, 4\)

so trying to understand the question from a different perspective would be quite useful to me.

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