# How would I find the big $\Theta$ of the following function?

$f(n) = \frac{n}{\log(n)}$

I understand the basics of how to find big O, Ω, and θ, however this particular function is giving me a lot of grief.

To be more clear, I will give a simple example of what I am looking for.

I know that the function $7n^5 - n^3 + n ∉ θ(n^5)$. I know this because $n^5$ has dominance over $n^3$ and over $n$ simply because the exponent is larger. I also know that when you add something that is of lower dominance to something of higher dominance, the complexity does not get altered. I can prove the inclusion of $7n^5 - n^3 + n$ in $θ(n^5)$ by taking the limit of $\frac{n^5}{7n^5 - n^3 + n}$ and confirming that the result is a constant.

The function $f(n)$ is not so simple. I do not know how to deal with division in this case, so I changed the function to only include what I know how to deal with. $f(n) = n*log(n)^-1$ (That is supposed to say "to the power of negative one".) From this point on, I have no clue what I should do. Please help?

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\Theta, please. –  Did Mar 4 '14 at 17:15
Use {} around exponents, so the whole thing is superscripted. –  Hurkyl Mar 4 '14 at 17:17

$$f(n) \in\Theta\left(\frac{n}{\log(n)}\right)$$