find a function $f:\mathbb{N}\to \mathbb{N}$ 
Find $f:\mathbb{N}\to\mathbb{N}$ s.t. $n^2 = o(f(n))$ and $f(n) = o(n^{2+\varepsilon})$.

So I thought about the function $f(n) = \left\lfloor n^{2+\varepsilon} \right\rfloor$
I already was able to prove that $f(n) = o(n^{2+\varepsilon})$, but had some troubles with the other part.
I tried to use the fact that $$\forall x. x-1 \le \lfloor x \rfloor$$
I'd be glad for help (by the way, is that a sufficiently elegant function for the problem?)
 A: $f(n) = n^2\ln(n)$.
$\frac{n^2}{f(n)}
=\frac1{\ln(n)}
\to 0$
so
$n^2 = o(f(n))$.
$\frac{f(n))}{n^{2+\epsilon}}
=\frac{\ln(n)}{n^{\epsilon}}
\to 0
$
since
$\ln(n) = o(n^{\epsilon})$
for any $\epsilon > 0$
so
$f(n)
= o(n^{2+\epsilon})
$
for any $\epsilon > 0$.
Note:
To show
$\frac{\ln(n)}{n^{\epsilon}}
\to 0$,
$\begin{array}\\
\frac{\ln(n)}{n^{\epsilon}}
&=\frac1{\epsilon}\frac{\epsilon\ln(n)}{n^{\epsilon}}\\
&=\frac1{\epsilon}\frac{ln(n^{\epsilon})}{n^{\epsilon}}\\
&\to 0
\end{array}
$
since
$\frac{\ln(n)}{n}
\to 0
$.
A: For $n>1$: Let $g(n)=1/\sqrt {\log n}$ .
Let $f(n)=$Floor$(n^{2+g(n)}).$
Then $f(n)n^{-2}\geq$  $(n^{2+g(n)}-1)n^{-2}=$ $n^{g(n)}-n^{-2}=$ $(e^{\log n})^{g(n)}-n^{-2}=$ $e^{\sqrt {\log n}}-n^{-2}$, which $\to \infty$ as $n\to \infty.$
And for any fixed $\epsilon >0$ we have $n^{2+\epsilon}f(n)^{-1}\geq$ $n^{2+\epsilon}n^{-2-g(n)}=$ $n^{\epsilon-1/\sqrt {\log n}}$, which $\to \infty$ as $n\to \infty.$
Remark: Floor$(n^{2+\epsilon})$ is NOT $o(n^{2+\epsilon})$ as $n\to \infty.$ when  $\epsilon >0.$
Because $1\geq$ Floor$(n^{2+\epsilon})n^{-2-\epsilon}\geq 1-n^{-2-\epsilon},$ which $\to 1$ as $n\to \infty.$
