Let $X=\mathbb{B}(0,1)$ denote the space of bounded functions from $(0, 1) \rightarrow \mathbb{R}$ with the supremum-metric. Does the sequence $(f_n)_{n=0}^\infty,f_n(x)=\frac{1}{nx}$ converge in $X$?
So we know that $X$ is complete and we can show that $f_n$ is Cauchy: Given $\epsilon >0$, we need $N\in\mathbb{N}$ such that $\forall m,n \in\mathbb{N}$ with $m,n>N$ , we have $sup_{x\in(0,1)}|\frac{1}{nx}-\frac{1}{mx}|<\epsilon \iff sup_{x\in(0,1)}(\frac{1}{|x|}|\frac{1}{n}-\frac{1}{m}|)<\epsilon\iff \frac{1}{|x|}sup_{x\in(0,1)}(|\frac{1}{n}-\frac{1}{m}|)=\frac{1}{|x|}(|\frac{1}{n}-\frac{1}{m}|)<\epsilon\iff (|\frac{1}{n}-\frac{1}{m}|)<\epsilon|x|$. Picking for simplicity $m=n+1$, we have $(|\frac{1}{n}-\frac{1}{n+1}|)=|\frac{1}{n(n+1)}|=\frac{1}{n(n+1)}<\frac{1}{n^2}<\epsilon|x|\iff n>\sqrt{\frac{1}{\epsilon|x|}}$ By similar calculations, we can compute the point-wise limit to be $0$. Is what I've done correct? Does there exist a point-wise limit that the function converges to?
Your calcualtions are not valid. You cannot pull out $\frac 1 {|x|}$ from the supremum over $x$.
For every fixed $x$ $\lim f_n(x)=0$ but the convergence in the supremum metric fails. This is because $\sup_x |f_n(x)-0|\geq |f_n(\frac1 n)| =1$.