How do I show $f_n(x)=n^2 x^n(1-x)$ pointwise converges to $0$ on $[0,1]$? 
How do I show $f_n(x)=n^2 x^n(1-x)$ pointwise converges to $0$ on $[0,1]$?


I first started with the case $x=1/2$ to try out. We see that $n^2(1/2)^n(1/2)=n^2/2^{n+1}$ which goes to $0$ as $n$ goes to infinity since $n^2 < 2^{n+1}$.
Now I need to generalize. If I take the derivative of $f_n(x)$ then I get, after simplifying,
$$f_n'(x)= n^2x^{n-1}(n-(n+1)x).$$
So, we know $x=0$ and $x=n/(n+1)$ are critical points.
I plotted the graph and I see that as $n$ gets larger, the maximum of the function on $[0,1]$ grows and gets closer to $x=1$.

HINT: Note that $n^2x^n(1-x) \to 0$, as $n \to \infty$ when $x \in [0,1]$, as the exponential term $x^n$ is the dominant term.

The easiest way would be to use the L'Hopital rule. Write $x = \frac 1y$, where $y > 1$. (The case when $x=1$ or $0$) can be considered separately and it's trivially true).  Then:
$$\lim_{n \to \infty} n^2x^n(1-x) = \lim_{n \to \infty} \frac{n^2(1-\frac 1y)}{y^n} = \lim_{n \to \infty} \frac{2n(1-\frac 1y)}{y^n\cdot \ln y} = \frac{2(1-\frac 1y)}{y^n \cdot (\ln y)^2} = 0$$
as the numerator is a constant and $\ln y > 0$
