Let $(f_n)^\infty_{n=1}$ be the seq. of functions $f_{n} : \mathbb{R} \rightarrow \mathbb{R}$ defined by $$f_n(x) = \left(\sin \left(\frac{x}{\pi}\right)\right)^{2n},~~~~\forall x \in \mathbb{R}$$
Find the pointwise limit of the sequence $(f_n)^\infty_{n=1}$ and does it converge uniformly.
So my thoughts were to do as how I would normally would do so, but with a certain limit of it all i.e sub in $x=0$ etc.
We have no limit here so I go to the second method - If i take the absolute value of $|f_n|$ and realise that $|\sin(x)| \leq 1 $ and then use it I just get
$$-1 \leq f_n(x) \leq 1$$
Which doesn't get me anywhere so I was hoping someone could help with this question thank you
It holds $\sin(\frac x \pi) = \pm 1$ if and only if $x= \pi(\frac \pi 2+ k\pi)$, for $k\in \mathbb Z$. So for $x_0= \pi(\frac \pi 2+ k\pi)$ you got $$\lim_{n\to \infty} \left(\sin \left(\dfrac {x_0} \pi \right)\right)^{2n}= \lim_{n\to \infty} (\pm 1)^{2n}=1.$$ For the other values of $x$ it holds $-1<\sin \left(\dfrac {x} \pi \right)<1$, hence $u:=\sin \left(\dfrac {x} \pi \right)^2\in[0,1)$. Therefore $$\lim_{n\to \infty} \left(\sin \left(\dfrac {x} \pi \right)\right)^{2n}= \lim_{n\to \infty} (u)^{n}=0.$$
The sequence converges pointwise to the discontinuous function $$f:\mathbb R \to \mathbb R\,, f(x)= \left\{ \begin{array}{ll} 1& \text{ if } x\in \frac{\pi^2} 2+ \pi^2\mathbb Z\\ 0& \text{ else.} \end{array}\right.$$ Since all the function $f_n$'s are continuous, but the limit $f$ is not, the convergence is not uniform.