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?

You do not have to apply L'hospital.
Let $x \neq 0$ or $1$ (trivial cases). Define $x_n=n^{2}x(1-x)^n$. We have
$$
x_{n+1}=\frac{(n+1)^2}{n^2}(1-x)x_n 
$$
implying ($x\in $(0,1)) that there is a value of $n$ after which the $x_{n+1}<x_n$. Furthermore $x_n>0$, hence it converges to $L$. Taking the limit on both sides of the equation we get
$$
L=(1-x)L
$$
leading to $L=0$.
