# Proving the uniform convergence of $f_n (x) = \sin(\frac{1}{n})\cos(x)$

I need to find the pointwise limit and determine the uniform convergence of the sequence of functions $$f_n (x) = \sin(\frac{1}{n})\cos(x)$$ where $$f$$ maps from reals to reals.

I got the pointwise limit to be $$0$$, but I am unsure how to prove the uniform convergence. I believe it starts with bounding this function -- upper bound is $$1$$ and lower bound is $$-1$$. I'm not sure where to go from there.

Hint: $$|\sin(x)| \leqslant |x|$$ So what can you say about $$||f_n-0||_u=\sup_x\left|\sin\left(\frac{1}{n}\right)\cos(x)\right|\text{ ?}$$
• Let $(E,d) , (E', d')$ be metric spaces. If given any $\epsilon > 0$, there is a positive integer $N$ such that $d'(f(p), f_n(p))<\epsilon$ whenever $n > N$ for all $p \in E$ – kt046172 Nov 19 '19 at 16:59
• @kt046172 I guess in your case both of the metric spaces are $\mathbb{R}$ with the euclidian metric. In that case, $|\sin(1/n)\cos(x)-0|\leqslant \frac{|\cos(x)|}{n}$. Can you continue? – Botond Nov 19 '19 at 17:56