Arrange the following growth rates in increasing order: $O (n (\log n)^2), O (35^n), O(35n^2 + 11), O(1), O(n \log n)$ I want to Arrange the following growth rates in increasing order
This order are following : $O (n (\log n)^2), O ((35)^n), O(35n^2 + 11), O(1), O(n \log n)$
Please give me idea how to arrange growth rates in increasing order
Regards,
Jatin
 A: If 
\begin{align}
\lim_{n \to \infty} \frac{f(n)}{g(n)} = \infty,
\end{align}
then the growth rate of $f(n)$ is larger than the growth rate of $g(n)$. Vice versa, if
\begin{align}
\lim_{n \to \infty} \frac{f(n)}{g(n)} = 0,
\end{align}
then the growth rate of $f(n)$ is smaller than the growth rate of $g(n)$.
Thus, the largest growth rate is $O(35^n)$, since
\begin{align}
\lim_{n \to \infty} \frac{35^n}{n \log^2 n} &= \infty, \\
\lim_{n \to \infty} \frac{35^n}{35n^2 + 11} &= \infty, \\
\lim_{n \to \infty} \frac{35^n}{1} &= \infty, \\
\lim_{n \to \infty} \frac{35^n}{n \log n} &= \infty.
\end{align}
And you can do the rest yourself.
A: When you are talking about big O notation, you should discard the constants, which means that $O(35n^2+11)$ should be written as $O(n^2)$. 
what you are interested in is the size at big numbers. And here is a rule to remember:
$$O(n^n)>O(n!)>O(\alpha^n)>O(n^\alpha)>O(log(n))>O(1)$$
Where $\alpha$ is a constant.
If you are not sure, then you can just insert a big $n$ and check which is the biggest ($n=100$):


*

*$35^{100} = 2.5\cdot10^{154}$

*$100^2 = 10^{4}$

*$100{\cdot}log^2(100)=400$

*$100{\cdot}log(100)=200$

*$1 = 1$ for every $n$
Showing you that $O(35^n) > O(n^2) > O(nlog^2(n))>O(nlog(n))>O(1)$
