Diameter of $A = \{f_n \mid n\in \mathbb{N} \}$ where $f_n(x)=x^n$. 
Let $C[0,1]$ be the set of continuous functions on $[0,1]$ with the sup-norm. Determine the diameter of the set $A = \{f_n \mid n\in \mathbb{N} \}$ where $f_n(x)=x^n$.

The diameter is clearly $1$. To show this I would need to show that $d(A) \leqslant 1$ and $d(A) \geqslant 1$. For the first case letting $f_n,g_m \in A$ I have that on $[0,1]$ the sup norm gives $$d(f,g) = ||f-g|| = \sup|f(x) -g(x)| = |x^n-x^m| \leqslant 1$$ thus $d(A) \leqslant 1$. However I'm not sure how to show the other way $d(A) \geqslant 1$, how should I approach this?
 A: Let $F(x)=x^m-x^n,$ where $x\in[0,1]$ ans suppose $n\gt m.$ Then $F'(x)=(m-nx^{n-m})x^{m-1}$ and therefore $F$ has its maximum (this need a little more than the first derivative) at $\left(\frac{m}{n}\right)^{1/(n-m)}.$ This gives an explicit expression the distance between $f_m$ and $f_n$ in your set as $$||f_m-f_n|| =\left(\dfrac{m}{n}\right)^{m/(n-m)}-\left(\dfrac{m}{n}\right)^{n/(n-m)}.$$ Now compute the limit of the distance $n\to\infty$ at $m=1$ to get what you want.
Added: I think here $m=1$ needs a bit more explanation. Note that distance between $f_n$ and $f_m$ gits larger as they gets apart. You can see this by plotting $n=1, 2, 3$ on the same graph (and $n=0,$ if you allowed it). So to make the maximum distance between to we can fix $m=1$ (or any other fixed natural also would work) and let $n\to\infty.$
A: Let $0\leq a<1$ and $a<x<1$. Since $x^n\to0$ as $n\to\infty$ there is some $n\in\Bbb N$ such that $x^n<x-a$, i.e. $x-x^n>a$. Hence $\|f_1-f_n\|>a$ and it follows that $d(A)\geq\sup_{n\in\Bbb N}\|f_1-f_n\|\geq 1$.
