Showing that $n\sin(x/n)$ does not converge uniformly on $\mathbb{R}$

Question

I know the version of this question with closed interval has been asked to death in this site, but how about if the function sequence $f_n(x) = n\sin(x/n)$ is considered over $\mathbb{R}$? At first I thought that the sequence would not converge to the limiting function $f(x) = x$, since at the points $x_n = n\pi$, $f(x_n) = n\pi$, but $f_n(x_n) = 0$. But is this sufficient to show that $\lim_{n \to \infty}\sup\{\left|f(x) - f_n(x)\right|\} \neq 0$?

Answer 1

Hint: In fact $$ \forall n \in \mathbb N:\sup_{x\in \mathbb R}\{|f(x) - f_n(x)|\} = \sup_{x\in \mathbb R}\{|x-n\sin(x/n)|\} =\ "\infty" $$ i.e. there is no supremum for any $n \in \mathbb N$ and so the limit $$ \lim_{n \to \infty}\sup\{\left|f(x) - f_n(x)\right|\} $$ is meaningless