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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.

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Hint: $$|\sin(x)| \leqslant |x|$$ So what can you say about $$||f_n-0||_u=\sup_x\left|\sin\left(\frac{1}{n}\right)\cos(x)\right|\text{ ?}$$

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  • $\begingroup$ What do the double absolute value and subscript u represent? $\endgroup$ – kt046172 Nov 19 '19 at 3:12
  • $\begingroup$ @kt046172 It's the uniform norm. How do you define the uniform convergence? $\endgroup$ – Botond Nov 19 '19 at 6:38
  • $\begingroup$ 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 $ $\endgroup$ – kt046172 Nov 19 '19 at 16:59
  • $\begingroup$ @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? $\endgroup$ – Botond Nov 19 '19 at 17:56
  • $\begingroup$ Sorry, where does the $\frac{cos(X)}{n} come from? I'm lost $\endgroup$ – kt046172 Nov 19 '19 at 18:28

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