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$$f(n)= 3n \log\log n$$

and $$g(n)=n (\log n)^6 + 8n^2$$ and I try to find

$$\lim_{n \rightarrow \infty} \frac{f(n)}{g(n)}$$

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    You mean $f(n)=3n\log\log n$ and $g(n)=n(\log n)^6 + 8n^2$? – 2017-01-30
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    @MPW Yes, I don't know how to write this porerly. Sorry about that. – 2017-01-30
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    No problem, just specifically wondered if the 6 was iteration or exponent. – 2017-01-30

2 Answers 2

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Hint:

$$\frac{n\log\log n}{n\log^6n+8n^2}\le\frac{n\log n}{8n^2}\le\frac{1}{n^{1/2}}$$

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    Thank you firstly. Then can we say it goes to 0 ? – 2017-01-30
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    @EminÇiftçi Indeed, using for example the squeeze theorem. – 2017-01-30
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    Oh, okey I see now clearly. Thank you again. – 2017-01-30
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Well, $g(n)\sim_\infty 8n^2$, hence $$\frac{f(n)}{g(n)}\sim_\infty\frac{3n \log\log n}{8n^2}=\frac38\frac{\log\log n}{n}\to 0,$$ as $\log\log n\in o(\log n)\in o(n)$.