sequence of functions converging uniformly on compact sets but not on the real line Consider the sequence of functions
$$
f_n(x)=n\sin((4\pi^2 n^2+x^2)^{1/2}), 0\leq x\leq C, C>0
$$
I cannot prove $(f_n)$ converges uniformly on $[0,C]$ by finding f(x) such that 
$$
\lim_n \sup_{0\leq x\leq C}||f_n(x)-f(x)||=0
$$
But is the sequence uniformly convergent in $\mathbb{R}?$
I think not, because it oscillates to infinity for large x. 
But how to prove it?
 A: Let's study pointwise convergence first. For fixed $x$ you have
$$\lim_{n \to \infty} 2 \pi n \left( \sqrt{1 + \frac{x^2}{(2 \pi n)^2}} -1\right) =  \lim_{n \to \infty} 2 \pi n \left( \frac{x^2}{2(2 \pi n)^2}\right) = 0 $$
where we used the asymptotic approximation $\sqrt{1+t^2} \sim \frac{t^2}{2}$. Hence
$$n \sin \left( \sqrt{(2 \pi n)^2 + x^2} \right) = n \sin \left( 2 \pi n  \sqrt{1 + \frac{x^2}{(2 \pi n)^2}} \right) = n \sin \left( 2 \pi n  \sqrt{1 + \frac{x^2}{(2 \pi n)^2}}  - 2 \pi n\right) =$$
$$= n \sin \left( 2 \pi n \left( \sqrt{1 + \frac{x^2}{(2 \pi n)^2}} -1 \right) \right)$$
which is of the form $\infty \cdot 0$ as $n \to \infty$. Using again the asymptotics above and $\sin t \sim t$, we can see that this limit equals
$$\lim_{n \to \infty} n \left( 2 \pi n \left( \frac{x^2}{2(2 \pi n)^2}\right) \right) = \frac{x^2}{4 \pi}$$
So we can define $f(x)= x^2/4 \pi$, and say that this is the pointwise limit of the $f_n$s.
Clearly $f$ is not bounded on $\Bbb R$, and all the $f_n$ have the trivial bound $\sup |f_n| \le n$, hence there is no uniform convergence on $\Bbb R$ (because for all $n$ we have $\sup |f_n - f| = \infty$).
A: I find this sequence a little bit akward because it seems to not simply converge. Could you provide a proof of uniform convergence ?
But if the question is about uniform convergence in $\mathbb{R}$, if you feel there's not, you can search for a particular point ( which can depend on $n$ ) that would show that the supposition is false. 
I suggest for example
$$ x_n=\sqrt{1-4\pi^2n^2}$$
Hence we have
$$\displaystyle f_n(x_n)=n \sin\left(1\right) \leq \underset{x \in \mathbb{R}}{\text{sup}}\left|f_n\left(x\right)\right|$$
But I still dont understand the simple convergence ( maybe i'm tired lol )
