# Showing uniform convergence of $f_n(x) = \sin(\frac{x}{n})$ on $[0, 1]$ and $[0, \infty]$.

I understand how to show that the sequence of functions given by $f_n(x) = \sin(\frac{x}{n})$ converges uniformly on $[0,1]$.

Indeed, suppose $\epsilon > 0$. Choose $N \in \mathbb{N}$ such that $N>\frac{1}{\epsilon}$. Then $\lvert \sin(\frac{x}{n}) \rvert \leq \frac{x}{n} \leq \frac{1}{n} \leq \epsilon$ if $n \geq N$.

I'm stuck on proving it fails to be uniformly convergent on $[0, \infty)$. I believe that to negate the definition of uniform convergence, I need to find an $\epsilon > 0$ such that for all $N \in \mathbb{N}$ there exists an $x \in [0, \infty)$ so we can find an $n \geq N$ such that $\lvert \sin(\frac{x}{n}) \rvert\geq \epsilon$.

So I guess could take $\epsilon = \frac{1}{10}$ and suppose $N \in \mathbb{N}$. But I don't know how to choose $x$ and find an $n\geq N$ to satisfy $\lvert \sin(\frac{x}{n}) \rvert\geq \epsilon$.

The sequence $\left(\sin\left(\frac xn\right)\right)_{n\in\mathbb N}$ doesn't converge uniformly on $[0,+\infty)$ because it converges pointwise to the null function, but, for each $n\in\mathbb N$, $\sin\left(\frac{\pi n}n\right)=-1$. So, if you take $\varepsilon=1$, it is not true that there is a natural $N$ such that$$n\geqslant N\wedge x\in[0,+\infty)\implies\left|\sin\left(\frac xn\right)\right|<\varepsilon.$$