Finding points of discontinuity of $f(x)=\lim_{t\to\infty}\frac{|a+\sin(\pi x)|^t-1}{|a+\sin(\pi x)|^t+1}$ I am studying continuity and came across the following function, whose continuity has to be determined, which is dependent on the values $a$ takes. 
$$f(x)=\lim_{t\to\infty}\frac{|a+\sin(\pi x)|^t-1}{|a+\sin(\pi x)|^t+1}$$
We can start off by making cases on the basis of when $|a+\sin(\pi x)|$ is less than $1$, exactly equal to $1$ and greater than $1$. This is a very tedious task. Is there a simpler way or a neat trick that can be employed to solve this problem. Thanks. 
 A: $f(x)=1-2\cdot\lim_{t\to\infty}\frac{1}{|a+\sin(\pi x)|^t+1}=
\begin{cases}
-1  & \text{$|a+\sin(\pi x)|<1$} \\
0 & \text{$|a+\sin(\pi x)|=1$}\\
1  &  \text{$|a+\sin(\pi x)|>1$}
\end{cases}$
Therefore, if $f$ is non-constant then $f$ does not have the intermediate value property and cannot be continuous. Furthermore, because $|a+\sin(\pi x)|$ is obviously not uniformly $1$ everywhere, $f$ is not the zero function. 
If $a\in[-2,2]$ then $a+\sin(\pi x_0)=\pm1$ for some real $x_0$, which implies $f$ is not continuous because $f(x_0)=0$. On the other hand, if $a\notin [-2,2]$ then $|a+\sin(\pi x)|>1$ for all $x$ implying $f$ is constant and thus continuous.
A: I think that essentially the cases must be made, but there are ways to be smart about them. If $U_a=\{x\in\Bbb R\,:\, \lvert a+\sin(\pi x)\rvert>1\}$ and $V_a=\{x\in\Bbb R\,:\, \lvert a+\sin(\pi x)\rvert<1\}$, then $f_a(x)=1$ for all $x\in U_a$ and $f_a(x)=-1$ for all $x\in V_a$. Since $U_a$ and $V_a$ are open, $f_a$ is continuous at all $x\in U_a\cup V_a$. Therefore the discontinuities are all in the set $$C_a=\{x\in\Bbb R\,:\, \lvert a+\sin(\pi x)\rvert=1\}=\Bbb R\setminus (U_a\cup V_a)$$ For all $x\in C_a$, $f_a(x)=0$, whereas $\lvert f_a(x)\rvert=1$ for all $x\notin C_a$. Therefore the only way for a point in $C_a$ not to be a discontinuity point is for it to be in the interior of $C_a$. However, it is quite clear that, for all $a\in\Bbb R$, $C_a$ is finite (possibly empty) union of sets of the form $\alpha+2\Bbb Z$, and therefore $C_a^\circ=\emptyset$.
Putting all together, the set of discontinuities of $f_a$ is $C_a$.
