Let $(f_n: \mathbb R\rightarrow \mathbb R)_{n\in\mathbb N}$ be the sequence of functions defined by
$$f_n(x)=\frac{x}{n}$$ for every $n\in\mathbb N$ and $x\in\mathbb R$. I would like to show $(f_n)_{n\in\mathbb N}$ does not converge uniformly on $\mathbb R$.
My attempt:
Let us suppose $(f_n)_{n\in\mathbb N}$ converges uniformly to $f: \mathbb R\longrightarrow \mathbb R$. In particular, $f_n(x)\to f(x)$ for every $x\in \mathbb R$ when $n\to \infty$. But:
$$0=\lim_{n\to \infty} \frac{x}{n}=\lim_{n\to \infty} f_n(x)=f(x)$$
for every $x\in\mathbb R$ and, therefore, $f=0$. Next, since $f_n\to f$ uniformly, given $\varepsilon=1$ there exists $n_0\in\mathbb N$ such that:
$$\forall n\geq n_0 \Rightarrow \frac{|x|}{n}=|f_n(x)|<1$$ for every $x\in\mathbb R$. But this is impossible since taking $n=n_0$ and $x=n_0+1$ we get
$$|f_n(x)|=\frac{|x|}{n}=\frac{n_0+1}{n_0}=1+\frac{1}{n_0}>1.$$
Is my proof correct? Is it standard or there is a more elegant way of writing?
Thanks.