Show that $f_n(x)=n^2x^n(1−x)$ converges pointwise to $0$ using epsilon argument from definition

Question

Let $f_n:[0,1]\to\Bbb R$ be defined by $f_n(x)=n^2x^n(1−x)$.

Show that $f_n(x)$ converges pointwise to $0$ using an epsilon argument from the definition of pointwise convergence.

I've got the following so far:

Let $\epsilon>0$. Then $N >$_______ implies:

$|n^2x^n(1-x)-0|=|n^2x^n(1-x)|\leq n^2x^n(1-x)$...

And I have no idea where to go from here, because I don't see how I can isolate the $n$ so that I can make the expression less than epsilon.

Thank you in advance!

Edit: I'm aware there's been a question posted about this before but the answers given have been expressed in terms of limits/theorems/tests/rules but I'd like to do it purely with an epsilon argument from the definition.

Answer 1

Let $x<1$ and $k=[\frac x {1-x}]+1$. Let $n_0>\frac {\log \epsilon} {\log [{(1+\frac 1 k) x}]}$ and $n_0 >1+2k^{2}$. Then $nx^{n}(1-x) <\epsilon$ for all $n \geq n_0$.

Sketch of proof: From the definition of $k$ we get $(1+\frac 1 k)x<1$. Also, the inequality $n_0 >1+2k^{2}$ shows that $n <\binom {n} {2} \frac 1{k^{2}}$ which implies $nx^{n}(1-x) \leq nx^{n} <((1+\frac 1 k )x)^{n}$. Finally we get $((1+\frac 1 k )x)^{n}<\epsilon$ from the fact that $n>\frac {\log \epsilon} {\log [{(1+\frac 1 k) x}]}$.