**Problem:**

prove that the sequence of functions $\displaystyle \left \{ f_{n} \right \}$ defined by:

$\displaystyle f_{n}\left ( x \right )= n\sin\sqrt{4\pi ^{2}n^{2}+x^{2}}$ converges uniformly on $\displaystyle \left [ 0,\alpha \right ]$ where $\displaystyle \alpha > 0$.

Does $\displaystyle \left \{ f_{n} \right \}$ converges uniformly on $\displaystyle \mathbb{R}$

Here is what I did: I proved that the pointwise limit of $\displaystyle \left \{ f_{n} \right \}$ is the function $\displaystyle f\left ( x \right )=\frac{x^{2}}{4\pi }$. Then, in order to prove the uniform convergence, I need to prove that $\displaystyle \sup_{x\in\left [ 0, \alpha \right ] } \left \{ \left | f_{n}\left ( x \right )-f\left ( x \right ) \right | \right \}n \to 0$ as $\displaystyle n \to \infty $ and that's where I am stuck. In the book, there is a hint saying that for x in the mentioned interval $\displaystyle x\in\left [ 0, \alpha \right ]$, and using the inequality $\displaystyle \sin\left ( x \right )\geq x-\frac{x^{3}}{3!}$,

we get:

$\displaystyle \left| n \sin\sqrt{4\pi ^{2}n^{2}+x^{2}} \right| -\frac{x^{2}}{4\pi }|\leqslant \frac{a^{2}}{4\pi }\left ( 1-\frac{2}{\sqrt{1+\frac{a^{2}}{4\pi ^{2}n^{2}}}+1} \right )+\frac{n}{3!}\frac{\alpha ^{6}}{8n^{3}\pi ^{3}} $

I don't understand how the book got this inequality based on $\displaystyle \sin\left ( x \right )\geq x-\frac{x^{3}}{3!}$. Can anyone give me a detailed proof how to get that inequality given by the book? From that point, I can easily prove the uniform convergence.

For the uniform convergence on R, there is hint saying that I should use $\displaystyle \left | \sin\left ( x \right ) \right | \leqslant \left | x \right |$ to get:

$\displaystyle \left| n \sin\sqrt{4\pi ^{2}n^{2}+x^{2}} \ -\frac{x^{2}}{4\pi }\right|\geq \frac{x^{2}}{4\pi }\left ( 1-\frac{2}{\sqrt{1+\frac{x^{2}}{4\pi ^{2}n^{2}}}+1} \right)$

Any help please how can we derive the last inequality too? because from this inequality, I can easily prove the non uniform convergence on R.