Limits: lim[n->infty] [cbrt{n^4+3n^7-2n^8} + cbrt{1-n^4+2n^8}]/[n^{1/6} sqrt{n^3-2}]

[EmanDroid]

How can I solve them?

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[math]\mbox{1. }\, \lim_{n \rightarrow \infty}\, \dfrac{\sqrt[3]{n^4\, +\, 3n^7\, -\, 2n^8\,}\, +\, \sqrt[3]{1\, -\, n^4\, +\, 2n^8\,}}{n^{(1/6)}\, \sqrt{n^3\, -\, 2\,}}[/math]
[math]\mbox{2. }\, \lim_{n \rightarrow \infty}\, \sum_0^n\, \left(\cos\left(\dfrac{2}{3}\pi\right)\right)^n[/math]

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Thank you

 

[Deleted member 4993]

How can I solve them?

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[math]\mbox{1. }\, \lim_{n \rightarrow \infty}\, \dfrac{\sqrt[3]{n^4\, +\, 3n^7\, -\, 2n^8\,}\, +\, \sqrt[3]{1\, -\, n^4\, +\, 2n^8\,}}{n^{(1/6)}\, \sqrt{n^3\, -\, 2\,}}[/math]
[math]\mbox{2. }\, \lim_{n \rightarrow \infty}\, \sum_0^n\, \left(\cos\left(\dfrac{2}{3}\pi\right)\right)^n[/math]

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Thank you

For problem #1, factor out n⁸ from n⁴ + 3n² -2n⁸ and do the same for 1 -n⁴ + 2n⁸.

What do you get?

 

[pka]

How can I solve them?

The second series is geometric: \(\displaystyle \mathop {\lim }\limits_{n \to \infty } {\sum\limits_{k = 0}^n {\cot \left( {\frac{{2\pi }}{3}} \right)} ^k}\)

Under what conditions does \(\displaystyle \sum\limits_{k = N}^\infty {{a_0}{r^k}} = \frac{{{a_0}{r^N}}}{{1 - r}}\;?\)