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 $x=1$.
 A: $x\in\{0,1\}$ makes the sequence constant. Otherwise $f_{n+1}(x)/f_n(x) = x\frac{n^2+2 n+1}{n^2}$. The fraction gets arbritrarily close to $1$ from above, so if $n$ is large enough, the product is less than $1$ and decreasing.
So if $n>N$ where $N$ is large enough, then $0\leq f_n(x)\leq a^{n-N}f_N(x)$ for some $a<1$. The RHS converge to $0$ exponentially.
A: For $0 < x < 1$ we have $y = 1/x - 1 > 0$ and $ x = 1/ (1 + (1/x - 1))= 1/(1+y).$
Thus,
$$n^2 x^n  = \frac{n^2}{(1 + y)^n}.$$
Using the binomial expansion $(1+y)^n = 1 + ny + \frac{1}{2}n(n-1)y^2 + \frac{1}{6}n(n-1)(n-2)y^3 + \ldots,$we have
$$0 \leqslant n^2 x^n  = \frac{n^2}{(1 + y)^n} < \frac{n^2}{n(n-1)(n-2)}\frac{6}{y^3},$$
and $\lim_{n \to \infty} n^2x^n = 0$ by the squeeze theorem.
A: 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 $\ln y > 0$
A: The case $x=0$ or $1$ are easy. 
For $x\in(0,1)$ we can find $\epsilon>0$ such that $ x < 1 - \epsilon$. 
Since $\frac{(n+1)^2}{n^2} \rightarrow 1$ we have that eventually 
$$\frac{f_{n+1}(x)}{f_n(x)} = \frac{x(n+1)^2}{n^2} < 1 -\frac{\epsilon}{2} $$ and so we are done by the ratio test.
