Show that $\{f_n\}_{n \geq 0}$ diverges in $(C[0,1], \|.\|)$. 
On $C[0,1]$, we consider the usual norm $\| f\|=\sup\{|f(t)|: t \in
[0,1]\}$. For $n\in N$, let $f_n \in C[0,1]$ define as
  $f_n(t)=t^{1/n}$. Show that $\{f_n\}_{n \geq 0}$ diverges in ($C[0,1]$,
  ||.||).

I thought I can use the contrapositive of the convergence of a sequence : $\exists \epsilon>0$ such that $\forall N\in\mathbb{N}$ $\exists n\ge N$ such that $|f_n-f|\ge \epsilon$. However, I don't know explicitly the function $f$.
Is anyone could give me a hint to solve the problem?
 A: I'm not sure why the pointwise argument is being abandoned in favor of the Cauchy argument. Suppose $f_n \to f$ in $C[0,1].$ This means $f_n \to f$ uniformly on $[0,1],$ which implies $f_n \to f$ pointwise on $[0,1].$ But as noted by others, $f_n(x) \to 1$ for $x\in (0,1]$ and $f_n(0) \to 0.$ Therefore $f=1$ on $(0,1]$ and $f(0)=0.$ That is impossible for a member of $C[0,1],$ hence we have a contradiction, proving that the $f_n$'s diverge.
A: Intuitively, the problem is that the pointwise limit has a jump at zero. But this is not a proof, and in fact there are examples (with a different norm) where the pointwise limit is discontinuous but you still have convergence to a continuous function (again, in a norm other than the uniform norm).
To actually give a proof, you should argue that the sequence is not Cauchy in the given norm: this means that
$$(\exists \varepsilon > 0)(\forall N \in \mathbb{N})(\exists m,n \geq N) \: \| f_n - f_m \| \geq \varepsilon.$$
Our intuitive observation gives us a hint about how to do this: we should look close to $t=0$. Accordingly, choose $\varepsilon=1/4$. Let $N \in \mathbb{N}$. Choose $m=N$. Find $t$ so that $t^{1/m}=1/4$. (What is this $t$?) Then find $n \geq N$ so that $t^{1/n} \geq 1/2$. (How can you find such a $n$?) Then $\| f_m - f_n \| \geq 1/4$ (why?).
A: Hint: You note that $t^{1/n}\rightarrow 1$ if $t\in(0,1]$ .
A: The "limit function," if it were to exist, would be $1$ on $(0, 1]$ and $0$ at $0$; this should give some intuition on how to show that the sequence is divergent. In particular, I'd suggest showing that this is not a Cauchy sequence, and hence is divergent; the behaviour of $t^{1/n}$ and $t^{1/(n + 1)}$ is different enough at $0$ to force divergence.
A: I get that for $n>m$, $\|f_n-f_m\|=(m/n)^{m/(n-m)}(1-m/n)$. (You'll want to check my calculation.) In particular, $\|f_{2m}-f_m\|=1/4$ for all $m$. This shows that $\{f_n\}$ can't be Cauchy.
