Calculation of the limit $\lim_{n \to +\infty} n^2x(1-x)^n, x \in [0,1]$ and the supremum How can I find this limit:
$$\lim_{n \to +\infty} n^2x(1-x)^n, x \in [0,1]$$
Do I have to use the L'Hospital rule? If so, do I have to differentiate with respect to $n$ or to $x$ ?
EDIT:
I also tried to find the supremum of $n^2x(1-x)^n$..I found that $f(x)=n^2x(1-x)^n$ achieves its maximum at the point $\frac{1}{n+1}$ and that $\sup_{x \in [0,1]} n^2x(1-x)^n=\frac{n}{e} \to +\infty$
Could you tell me if it is right?

This limit is always zero, there are two cases:
1)$x=0$-it's obvious.
2)$x \neq 0$, then the limit is zero, because $1-x <1$ and exponential function decrease faster than polynomial $n^2$
