TITLE: Showing that $\sin(\sqrt{4 \pi^{2}n^{2} + x})$ converges uniformly on $[0,1]$
QUESTION [5 upvotes]: Suppose we are considering the sequence of functions $f_{n}(x)=\sin(\sqrt{4 \pi^{2}n^{2} + x})$ and I am having trouble showing that that $f_{n}$ converges uniformly on the interval $[0,1]$. 

An idea, I've tried is to consider  the Taylor series:
$$\sin(\sqrt{4 \pi^{2}n^{2} + x}) = (\sqrt{4 \pi^{2}n^{2} + x})- \frac{(\sqrt{4 \pi^{2}n^{2} + x})^{3}}{6} + O((\sqrt{4 \pi^{2}n^{2} + x})^{5})$$
but I haven't gotten anything useful as of yet.

REPLY [3 votes]: Your sequence converges uniformly to $0$.
\begin{align}
\left|\sin\left(\sqrt{4\pi^2n^2+x}\right)\right| &= \left|\sin\left(\sqrt{4\pi^2n^2+x}\right)-\sin(2\pi n)\right| \\
&= \left|2\sin\left(\frac{\sqrt{4\pi^2n^2+x} - 2\pi n}2\right)\cos\left(\frac{\sqrt{4\pi^2n^2+x} + 2\pi n}2\right)\right|\\
&\le 2\left|\sin\left(\frac{\sqrt{4\pi^2n^2+x} - 2\pi n}2\right)\right|\\
&\le 2\left|\frac{\sqrt{4\pi^2n^2+x} - 2\pi n}2\right|\\
&= \sqrt{4\pi^2n^2+x} - 2\pi n\\
&= \frac{4\pi^2n^2+x - 4\pi^2n^2}{\sqrt{4\pi^2n^2+x} + 2\pi n}\\
&= \frac{x}{\sqrt{4\pi^2n^2+x} + 2\pi n}\\
&\le \frac{1}{4\pi n} \xrightarrow{n\to\infty} 0
\end{align}
uniformly in $x \in [0,1]$.
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