Let $f_n:[0,2] \to \mathbb{R}$, $f_n=\cos(\pi+\frac{x}{n})$. Show the sequence $f_n$ converges uniformly, and find the uniform limit.
I often struggle with proving uniform convergence, I wasn't sure if I should try to do it from first principles, or use the fact that $[0,2]$ is compact, and try to show that $f_n(x)$ is decreasing after some $n$, and then use Dini's theorem. I dont seem to be able to show $f_n$ is decreasing at each $x$ without pictures. Thanks
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Technically, everything is said in the comments, but let's emphasize the concept, because it seems to be difficult for many people: usually, it's simple to see the pointwise limit $f(x)$ of a function sequence $f_n(x)$, in this case, that's just the constant $\cos\pi=-1.$ The "uniform" now just means to find an estimate for the difference $|f_n(x)-f(x)|$ valid for all $x\in[0,2]$, i.e. it does not depend on $x$ (and has to converge to $0$ as $n\rightarrow\infty,$ naturally). In this case, you can use $$\left|\cos(\pi+\frac{x}{n}-\cos\pi\right|\le\frac{x}{n}\le\frac{2}{n},$$ because the derivative of $\cos$ is $\le1$ everywhere.
It's the latter estimate, $\frac{2}{n}$ independent of $x$, that turns the trick, and that's possible only because the interval is finite. On the whole real axis, there is convergence in this case as well, but no uniform convergence.
First, note that if a uniform limit exists, it must coincide with the pointwise limit. At any $x$, we calculate $$ \lim_{n \to \infty} \cos\left(\pi + \frac xn\right) = \cos \pi = -1 $$ Now, we wish to show that the convergence is uniform. That is, we wish to find a uniform upper bound for the difference $|\cos(\pi) - \cos(\pi + x/n)|$. A handy trick here is to use the mean value theorem on the function $x \mapsto \cos(\pi + x/n)$: for all $x \in [0,2]$, there exists a $c \in (0,x)$ such that $$ (\cos(\pi) - \cos(\pi + x/n)) = \left(\left.\frac d{dx} \cos(\pi + x/n)\right|_{x=c} \right) \cdot x $$ perhaps you could take it from there.