Let $f_n \in C([0,+\infty))$ be defined by $f_n(t):=\sin{\sqrt{t+4n^2\pi^2}}$, for $n \in \mathbb N$ and $t \ge 0$.
Prove that $f_n$ converges pointwise to $f \in C([0,+\infty))$ and determine $f$;
study the uniform convergence of the sequence on bounded intervals and on $[0;+\infty)$.
Well, I've got some problems and I need your kind help.
First of all, I've noted that $f_n(0)=0$ for every $n \in \mathbb N$. My guess is that the pointwise limit is $f \equiv 0$. But how can I prove it rigorously? I think that one should note that $4n^2\pi^2=(2n\pi)^2$ so there must be something related to periodicity of the function $\sin(\cdot)$...
Thanks in advance.