Proof that $\frac{x}{1+nx}$ converges uniformly to the zero function on $[0,1]$

Question

I am trying to prove that $f_{n}(x)=\frac{x}{1+nx}$ converges uniformly to the zero function on $[0,1]$. Is the following proof correct?

Consider some arbitrary $\epsilon>0$. Set $N\triangleq \frac{1-\epsilon}{\epsilon }$. Consider an arbitrary $n>N$ and $x\in[0,1]$. Then we have that $n>\frac{1-\epsilon}{\epsilon}>\frac{x-\epsilon}{\epsilon x}$ (where the last inequality follows from the fact that: $\sup \{\frac{x-\epsilon}{\epsilon x}:x\in[0,1], \epsilon>0\}\triangleq\frac{1-\epsilon}{\epsilon}$, since $\frac{x-\epsilon}{\epsilon x}$ can be shown to be monotone increasing on $[0,1]$). So $n\epsilon x>x-\epsilon \iff \frac{x}{nx+1}<\epsilon$.

So I have shown that, given some arbitrary $\epsilon>0$, there exists some $N\in \mathbb{N}$ s.t. for all $n>N$ and $x\in[0,1]$, it is the case that $|\frac{x}{1+nx}|<\epsilon$

Answer 1

That's fine, but it can be made much easier: $$|\frac{x}{1+nx}|=\frac{x}{1+nx}\le \frac{x}{nx}=\frac{1}{n}\rightarrow 0$$

Answer 2

We have $$ f(0)=0<\frac1n \quad \forall n\ge 1, $$ and for $x > 0$ we have $$ f_n(x) < \frac{x}{nx}=\frac1n \quad \forall n\ge 1. $$ It follows that $$ |f_n(x)|=f_n(x)< \frac1n \quad x\ge 0, n \ge 1. $$

Hence $\displaystyle \|f_n\|_{\infty}=\max_{0\le x \le 1}|f_n(x)|<\frac1n $ and $\displaystyle \lim_{n}\|f_n\|_{\infty}=0$.