Difficulty with limit to show pointwise convergence (without L'hospital's rule) I've come across a question which asks to show pointwise convergence on $[0,1]$ for the following sequence of functions defined for integers $n\geq0$ :  $f_n(x)=n^2x^n(1-x)$. I know that the sequence converges to 0 for all $x\in[0,1]$ but I'm having trouble proving these limits exist and equal 0 without L'hospital's rule.
For $x=0$ and $x=1$ it's clear that $f_n(x)=0$ so $\lim_\limits{n\to\infty} f_n(x)= 0 $ but for $c\in(0,1)$ we have  $f_n(c)=n^2c^n(1-c)$ which I'm finding harder to prove neatly. One approach I took is $$\lim_\limits{n\to\infty}f_n(c)=\lim_\limits{n\to\infty}n^2c^n(1-c)=$$$$(1-c)\lim_\limits{n\to\infty}\frac{n^2}{c^{-n}}=(1-c)\lim_\limits{n\to\infty}\frac{2n}{-nc^{-n-1}}= $$$$(c-1)\lim_\limits{n\to\infty}\frac{2}{c^{-n-1}}= 2(c-1)\lim_\limits{n\to\infty}c^{n+1}=0$$
My problem here is that since the limit exists we should be able to choose an $N$ for a give $\epsilon>0$ such that $n>N \implies n^2x^n(1-x)<\epsilon$ but I'm having diffuclty choosing such an $N$. I know that since $$\int_0^1f_n(x)dx=1\neq\int_0^1f(x)dx=0$$, the sequence does not congverge uniformly and so our choice of $N$ must depend on the specific $x\in[0,1]$. 
So far I've tried simplifying the limit by saying that since $n^2x^n(1-x)<n^2x^n$ for $x \in (0,1)$ it will suffice to to choose $N$ such that $n>N \implies n^2x^n<\epsilon$. I feel like it'll be something in the form $N=\sqrt\frac{\epsilon}{2x^?}$ but I'm completely stuck. 
TLDR: How to prove $\lim_\limits{n\to\infty}n^2c^n=0$ for $c\in(0,1)$ from definition of a limit/without L'hospital's rule. 
 A: Since$$\lim_{n\to\infty}\left|\frac{(n+1)^2c^{n+1}}{n^2c^n}\right|=c,$$if you fix some $d\in(c,1)$, there is a $p\in\mathbb N$ such that $n\geqslant p\implies\left|\frac{(n+1)^2c^{n+1}}{n^2c^n}\right|\leqslant d$. So,


*

*$(p+1)^2c^{p+1}\leqslant dp^2c^p$;

*$(p+2)^2c^{p+2}\leqslant d^2p^2c^p$

*$\vdots$

*$(p+k)^2c^{p+k}\leqslant d^kp^2c^p$.


So, by the squeeze theorem, $\lim_{n\to\infty}n^2c^n=0$.
A: OK, let's try this. Since $0<c<1$, $\dfrac1c = 1+\delta$ for some $\delta>0$. By the binomial theorem, for $n\ge 3$ we have $(1+\delta)^n \ge 1+n\delta+\dfrac{n(n-1)}2\delta^2+\dfrac{n(n-1)(n-2)}6\delta^3>\dfrac{n(n-1)(n-2)}6\delta^3$. Thus, $$n^2c^n = \frac{n^2}{(1+\delta)^n}<\frac{n^2}{\frac{n(n-1)(n-2)}6\delta^3}\le\frac{24n^2}{n^3\delta^3} \text{ if } n\ge 4$$ (as this guarantees $n-1>n-2\ge 2$). Finally, if $n\ge 4$, we have $$n^2c^n < \frac{24}{n\delta^3},$$
which in turn will be less than a given $\varepsilon>0$ provided $n>\max\left(3,\dfrac{24}{\varepsilon\delta^3}\right)$.
A: You want to prove that, for $0<x<1$,
$$
\lim_{n\to\infty}n^2x^n=0
$$
as the convergence to zero of $(n^2x(1-x))_n$ is obvious for $x=0$ and $x=1$.
This in turn is equivalent to prove that, for $y>1$,
$$
\lim_{n\to\infty}\frac{n^2}{y^n}=0
$$
or as well that
$$
\lim_{n\to\infty}\frac{y^n}{n^2}=\infty
$$
If $z=\sqrt{y}>1$, we just need to show that
$$
\lim_{n\to\infty}\frac{z^n}{n}=\infty
$$
Write $z=1+t$, with $t>0$; by the binomial theorem
$$
z^n=(1+t)^2>1+nt+\frac{n(n-1)}{2}t^2
$$
because we omit positive terms in the binomial expansion. Hence
$$
\frac{z^n}{n}>\frac{1}{n}+t+\frac{(n-1)t^2}{2}
$$
and the right-hand side has limit $\infty$.
Note that a similar argument shows
$$
\lim_{n\to\infty}n^kx^n(1-x)=0
$$
for every $x\in[0,1]$ and every positive integer $k$, just use $z=\sqrt[k]{1/x}$.
A: I'm just going to add this: Exponential decay always wins over polynomial growth. Always. So if $0<x<1,$ then $x^n\to 0$ exponentially fast. If you try to subvert this by mulitplying $x^n$ by $n^p$ for some large $p,$ you will fail; it will still go to $0.$
You need to feel this in your bones as you progress in your mathematics studies.
And PS: L'Hopital is not a good way to see this. This is, essentially, a pre-calculus result.
