Discontinuities of a Function Find the set of discontinuities of:
$$f(x)= \lim_{n \to\infty} \left(\lim_{t\to 0} \frac{(sin(n!\pi x))^2}{(sin(n!\pi x))^2+t^2}\right)$$
I was thinking of using sterlings formula to simplify the $n!$ part, but the fact that $t$ tends to $0$ has confused me. Is it that $n$ can take only integer values?
 A: While, without specifying whether $n \in \mathbb{N}$ or $n \in \mathbb{R}$, we do not know if $n$ can take only integer values.
However, judging on the $n!$, I assume that $n \in \mathbb{N}$. And otherwise, the second limit is non-determined.
Now, we have can have $\sin( n! \pi x) = 0$ or $\sin(n! \pi x) \neq 0$.
In the first case, we have
$$\lim_{t \to 0} \frac{\sin( n! \pi x)^2}{\sin( n! \pi x)^2+t^2}= \lim_{t \to 0} \frac{ 0}{0+t^2} = \lim_{t \to 0} 0 = 0 = \sin( n! \pi x) $$
In the second case we have 
$$\lim_{t \to 0} \frac{\sin( n! \pi x)^2}{\sin( n! \pi x)^2+t^2}= \lim_{t \to 0} \frac{ \alpha}{\alpha+t^2} = \frac{\alpha}{\alpha} = 1$$
So we got rid of one limit.
Now, let $x \in \mathbb{Q}$, then $x = \frac{p}{q}$ for $p \in \mathbb{Z}$ and $q \in \mathbb{N}$. So in particular for all $n \ge q$, we have $n!x \in \mathbb{Z}$.
As we know, $\sin ( m \pi x) = 0$ for $m \in \mathbb{Z}$, so we have $f(x) = \lim_{n \to \infty} 0 = 0$.
Now, let $x \in \mathbb{R} \backslash \mathbb{Q}$, then we have $n!x \in \mathbb{R} \backslash \mathbb{Q}$ for all $n \in \mathbb{N}$. So we have $\sin( n! x \pi) \neq 0$ for all $n \in \mathbb{N}$. So $f(x) = \lim_{n\to\infty }1=1$.
So the set of discontinuities is $\mathbb{R}$.
A: Yes, variables named $n$ are often implicitly assumed only to range over integers, and this is almost certainly the case here.
Don't use Stirling's approximation here -- it's only an approximation, which is useless in this context because the important thing about the argument to the sine is not its size, but where it falls modulo $2\pi$.
Instead the point in constructions like these is usually that $n!\pi x$ will eventually be a multiple of $\pi$ when $x$ is rational, which means that eventually the sines are all $0$.
On the other hand, if $x$ is irrational, then $n!\pi x$ is never a multiple of $\pi$, and the sines are never $0$...
