<s>[INST] Uniform convergence of $n\sin\sqrt{4\pi^2n^2+x^2}$ on $[0,a]$ and $\mathbb{R}.$ Let $f_n(x)=n\sin\sqrt{4\pi^2n^2+x^2}$. Prove that $(f_n)$ is uniformly convergent on $[0,a]$ for every $a>0.$ Is the convergence uniform on $\mathbb{R}$?
Attempt.  For $x\in \mathbb{R}$ constant we have:
$$f_n(x)=n\sin(\sqrt{4\pi^2n^2+x^2}-2\pi n)=n\sin\bigg(\frac{x^2}{\sqrt{4\pi^2n^2+x^2}+2\pi n}\bigg)=
\frac{\sin\bigg(\frac{x^2}{\sqrt{4\pi^2n^2+x^2}+2\pi n}\bigg)}{\frac{x^2}{\sqrt{4\pi^2n^2+x^2}+2\pi n}}\,\frac{n\,x^2}{\sqrt{4\pi^2n^2+x^2}+2\pi n}\rightarrow 1\cdot \frac{x^2}{4\pi}=\frac{x^2}{4\pi},$$
so we are done with the pointwise convergence.
So far I am stuck on the uniform convergence, in other words how to prove $$\max_{0\leq  x\leq a}|f_n(x)-x^2/(4\pi)|\rightarrow 0$$ and $$\sup_{x\in \mathbb{R}}|f_n(x)-x^2/(4\pi)|=\sup_{x\geq 0}|f_n(x)-x^2/(4\pi)|\nrightarrow 0.$$
Thanks for the help.
 [/INST] I want to offer you two different ways… choose which you prefer:
1.) Take $y \ge 0$ s.t. $$y^2 = 2\pi^2n^2 + x^2$$
Then $$x \ge 0 \iff y \ge 2\pi n$$
And we get: 
$$\begin{align*}\sup_{x\geq 0}\left|f_n(x)-\frac{x^2}{4\pi}\right| &= \sup_{y \ge 2\pi n} \left|n\sin(y) - \frac{y^2 - 4\pi^2n^2}{4\pi}\right| \\\\&\ge \sup_{\substack{y = 2kn\pi \\ k\in \Bbb N}} \frac{4k^2n^2\pi^2 - 4\pi^2n^2}{4\pi} \\\\ &=\sup_{\substack{y = 2kn\pi \\ k\in \Bbb N}}(k^2-1)\pi n^2 \\ &= \infty\end{align*}$$
2.) Another way: We have $$-n \le f_n(x) \le n$$ and so $$-n -\frac{x^2}{4\pi} \le f_n(x)-\frac{x^2}{4\pi} \le n -\frac{x^2}{4\pi}$$
Hence we can conclude $$\lim_{x\to \infty} f_n(x) -\frac{x^2}{4\pi} = -\infty$$
And so we get directly: $$\sup_{x\geq 0}\left|f_n(x)-\frac{x^2}{4\pi}\right| = + \infty$$
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