Show Uniform Convergence of $f_n(x)=n^2x^n(1-x)^2$

Question

$f_n(x)=n^2x^n(1-x)^2$ on $[0,a]$ where $a<1$

I know that $f_n(x)$ converges to $f=0$

Uniformly converge iff: for all $\epsilon >0$ there exists $n>N$ s.t. $|f_n-f|<\epsilon$

So I should show that I can find $n$ such that $|f_n-f|=|n^2x^n(1-x)^2-0|<\epsilon$ for all $0\leq x \leq a$

I already showed that it is not uniformly convergent on [0,1] by taking $f_n(1-1/n)$ because $(1-1/n)^n$ is 1 as $n \rightarrow \infty$

Answer 1

It is bounded on $[0,a]$ by $n^2a^n$, which converges to $0$ without depending on $x$.

Answer 2

A hint is to analyze the functions $f_n(x)$ on the interval $[0,1]$, while keeping in mind that you really only care about what happens on $[0,a]$. Find the derivative, it's zeroes and try to graph the function. It might be useful to start with say $f_3(x)$, and then consider the general case $f_n(x)$ afterwards.

Knowing the shape and properties of the sequence of functions can go a long way in "guessing" an $n_\epsilon$ for a given $\epsilon > 0$.