Let $x\in[0,1]$, and define $f=x(1-\frac{1}{2}x)^n$, where $n\in\mathbb{N}$. Certainly $x(1-\frac{1}{2}x)^n\rightarrow 0$ as $n\rightarrow\infty$. How to prove that this convergence is uniform by it is bounded by $\frac{2}{n+1}$?
Attempt: The limit function is $0$, I tried to consider the maximum by using the first derivative. After that I have $nx=2-x$, which I'm having trouble to write $x$ in terms of $n$. Prove $x(1-\frac{1}{2}x)^n\rightarrow 0$ uniformly
As an alternative to finding the maximum, use the following estimate.
With $0 < x \leqslant 1$ we have $0 < (1 - x/2) < 1$. There exists $a > 0$ such that $(1 - x/2) = 1/(1 +a)$. In this case $a = x /(2-x)$ and, using Bernoulli's inequality $(1 +a)^n > na$, we have
$$x\left(1 - \frac{x}{2} \right)^n = \frac{x}{(1 + a)^n} < \frac{x}{na} .$$
From here you should be able to finish and show that $2/n$ is an upper bound.