Give two functions that is in O(n^2) but not in o(n^2).
The answer is nlogn and n,But I think a lot but can't understand why answer are them.I doubt answers is wrong.
And this question,Give two functions that is in big-omega(n^2) not in little-omega(n^2)?
The answer is n^2logn and n^3.But I think a lot but can't understand why answer are them.
Where have you got those answers?
1. Answers are wrong if that was the task. The simplest answer is $n^2$. Why? Because this function satisfies restructions: $\lim_{n\to\infty}\frac{f(x)}{n^2}>0$ (not in $o(n^2)$ requirement) and $\lim_{n\to\infty}\frac{n^2}{f(x)}>0$ (in $O(n^2)$ requirement).
2. Exactly the same answer.
The answers you give are wrong. It is easy to verify that $n, n\log n \in o(n^2)$, and similarly $n^2 \log n, n^3 \in ω(n^2)$.
Give two functions that is in $O(n^2)$ but not in $o(n^2)$.
Looking at the definitions, we see that there are two ways to come up with such a function:
Pick any function from $\Theta(n^2)$.
Besides the obvious candidates of the form $cn^2 + o(n^2)$, there are also more obscure ones like, for instance $ 2^{\frac{\sin n}{\log n}} \cdot n^2$.
Pick any function $f \in O(n^2)$ for which $\lim_{n \to \infty} f(n) \cdot n^{-2}$ does not exist.
Simple examples are $(2 + \sin(n)) \cdot n^2$ or $2^{\sin n} \cdot n^2$ or
$\qquad \displaystyle f(n) = \begin{cases}n^2, &n \in 2\mathbb{N};\\n, &n \in 2\mathbb{N}+1.\end{cases}$
There are many more.
I'll leave the proofs to you as an exercise, as well as the symmetric case with $\Omega$ vs $\omega$.
