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?

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 $\sup |f_n - f| = \infty$).
