For $k\geq 1$ we have the function $f_k : x \rightarrow > sin(\frac{x}{k})$. Show that $f_k \rightarrow 0$ uniform on $[-R,R]$ for every $R>0$.
To show it is uniform convergent I need to show there is some function $g$, such that $\lim_{k\rightarrow \infty} ||f_k-g||=0$.
My attempt was $$|f_k-g|=|sin(\frac{x}{k})-g| \leq |\frac{x}{k}-g|$$ So if we take $g=\frac{x-1}{k}$, we get $$=|\frac{1}{k}|$$ which goes to $0$ if $k$ goes to infinity.
I'm not sure if this is the right way to go, and if I'm even allowed to take $|sin(\frac{x}{k})-g| \leq |\frac{x}{k}-g|$ this step, since we don't know anything about $x$.
$|\sin (\frac x k)| \leq \frac {|x|} k \leq \frac R k$ which shows that $\sin (\frac x k) \to 0$ uniformly.
[By MVT $|\sin t|=|\sin t-\sin 0|=|t| |\cos s|$ for some $s$ so $|\sin t| \leq |t|$].