<s>[INST] Showing that $\sin(\sqrt{4 \pi^{2}n^{2} + x})$ converges uniformly on $[0,1]$ 
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.
 [/INST] Fix $x\in[0,1]$. The Mean Value Theorem  shows that $$|\sqrt{4\pi^2n^2+x}-2\pi n|<\frac x{4\pi n}\le\frac1{4\pi n}.$$ So $2\pi n \le \sqrt{4\pi^2n^2+x} < 2\pi n +1/(4\pi n)$. Then apply MVT again to show$$|\sin(\sqrt{4\pi^2n^2+x})|
=|\sin(\sqrt{4\pi^2n^2+x})-\sin(2\pi n)|<\frac1{4\pi n}.$$
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