contradiction on completness of $C^0_\infty$

Question

I am confused by the next problem:

Show that the sequence

$$f_n = \begin{cases} 0 & 0\leqslant x<2-\dfrac{1}{n} \\ 1+n(x-2) & 2-\dfrac{1}{n}\leqslant x<2 \\ 1 & 2\leqslant x\leqslant 3 \end{cases}$$

1. is a Cauchy sequence.
2. find the limit
3. it is $C^{0}_{\infty}[0,3]$ (the space of continuous functions with domain $[0,3]$ a Banach space?

My confusion is that the limit is actually a non-continuous function, so it is a counterexample to say that $C$ is not a Banach space. However, I already know that it is a Banach space because $[0,3]$ is compact. Therefore, I am suspicious that it is not a Cauchy sequence but I don't know how to prove it, so any help?

the limit i get was

$$f_n = \begin{cases} 0 & 0\leqslant x<2 \\ 1 & 2\leqslant x \leqslant 3 \end{cases}$$

Answer 1

Assume $m>n$. Then, since $2-\frac1n<2-\frac1{n+k}$, \begin{align} f_{n+k}(2-\frac1{n+k})-f_n(2-\frac1{n+k})&=-f_n(2-\frac1{n+k})=-\left(1+n\left(2-\frac1{n+k}-2\right)\right)\\ \ \\ &=-\left(1-\frac n{n+k}\right). \end{align} So, for instance, if $k>n$, then $$ \|f_{n+k}-f_n\|\geq\frac12. $$ So the sequence is not Cauchy.