# About the convergence of $f_n:=\sin{\sqrt{t+4n^2\pi^2}}$

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 $t \ge 0$.

1. Prove that $f_n$ converges pointwise to $f \in C([0,+\infty))$ and determine $f$;

2. 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 $\sin(\cdot)$...

Hint $$\sin \left( {\sqrt {t + 4{n^2}{\pi ^2}} } \right) = \sin \left( {2n\pi \sqrt {\frac{{t + 4{n^2}{\pi ^2}}}{{4{n^2}\pi }}} } \right) = \sin \left( {2n\pi \sqrt {\frac{t}{{4{n^2}\pi }} + 1} } \right)$$

and

$$\sqrt {1 + x} = 1 + \frac{x}{2} + o\left( x \right)$$ for $x\to 0$

Of course $\sin x =x+o(x)$ for $x\to 0$

$\sin(\sqrt{t+4n^2\pi^2})=\sin\left(2n\pi\sqrt{1+\frac{t}{4n^2\pi^2}}\right)\approx \sin\left(2n\pi+\frac{t}{4n\pi}\right)=\sin\left(\frac{t}{4n\pi}\right)\approx\frac{t}{4n\pi}\rightarrow 0$

so $f(t)=0$.

On any interval, the convergence is bounded as $|\sin(x)|\le|x|$.