How to prove that $f_n(x):=x^n$ is not a Cauchy sequence in $C[0,1]$ under the norm $\|f\|= \sup|f(x)|$? 
How to prove that $f_n(x)=x^n$ is not a Cauchy sequence in $C[0,1]$ under the norm $\|f\|= \sup_{x\in [a,b]}|f(x)|$, by showing that it does not satisfy the definition of a Cauchy sequence?


You want to show that
$$
\forall\epsilon>0,\exists N,\forall n,m:n,m>N\implies\|x^n-x^m\|=\max_{x\in[0,1]}|x^n-x^m|<\epsilon\tag{$\star$}
$$
is not satisfied.

Proposition: Every $\epsilon\in(0,1)$ fails to satisfy $(\star)$.

Proof: Choose any $\epsilon\in(0,1)$, $N\in\mathbb{N}$ and $m>N$. Let $x_0\in(0,1)$ be such that
$$
x_0^m=\epsilon+\frac{1-\epsilon}{2}
$$
Now, since $x_0^n\to0$ as $n\to\infty$, we see that upon choosing $n>m$ large enough we obtain
\begin{align}
|x_0^n-x_0^m| &= x_0^m-x_0^n\\
&=\epsilon+\frac{1-\epsilon}{2}-x_0^n\\
&>\epsilon
\end{align}
Remark: The idea is that for fixed $x_0$ and $m$, the value $x_0^m-x_0^n$ tends to $x_0^m$. Hence all we needed was to make sure that $x_0^m>\epsilon$.
