Uniform convergence How can I prove the sequence of functions $x(1-x), x^2(1-x), ...$ converges uniformly on $[0,1]$?
I know from my earlier question here that I can prove it by taking the derivative. But how can I prove it without taking the derivative since we still haven't defined "derivative" yet in my class? I know I can divide the regions into two pieces. 
 A: observe that the $x^n$ parts are always smaller or equal to $1$. Now take any $\epsilon > 0$, and consider the regions $[0, 1-\epsilon]$, $[1-\epsilon, 1]$. On the second one all your functions are smaller than $\epsilon$ thanks to the $1-x$ factor. to take care of the first one just take $n$ large enough such that $(1-\epsilon)^n < \epsilon$ and since $1-\epsilon$ is also smaller than $1$ you're done
A: Let $f_n(x) = x^n(1-x)$.
Note that $f_n(0) = f_n(1) = 0$ and $f_n (x) \ge 0$ for $x \in [0,1]$. Hence $f$ has a maximum in $(0,1)$.
Setting $f_n'(x) = 0 $ gives $x = \frac{n}{n+1} $, hence we have $f_n(x) \le \frac{\left( \frac{n}{n+1} \right)^n}{n+1} \le \frac{1}{1+n}$.
Hence $f_n \to 0$ uniformly on $[0,1]$.
Alternatively: Let $\epsilon>0$ and choose $\theta \in (0,1)$ such that $1-\theta < \epsilon$. Now choose $N$ such that $\theta^N < \epsilon$. Then for all $n \ge N$, we have $|f_n(x)| < \epsilon.$
To see this, if $x \in [0,\theta]$, we have $|f_n(x)| \le x^n \le \theta^n \le \theta^N < \epsilon$. If $x \in (\theta,1]$, $|f_n(x)| \le 1-x < 1 - \theta < \epsilon$.
