Find the points of discontinuities of the function Find the points of discontinuities of the function
$$\lim_{n\to \infty}\frac{(1+\sin{\frac{\pi}{x}})^n-1}{(1+\sin{\frac{\pi}{x}})^n+1}$$ $x\in(0,1)$.
Same question asked by me in Find the point of discontinuity of the two functions 1. $f(x)=[\sin{x}]$, [] std for greatest integer function
I don't know how to solve it. It is true that $\frac{(1+\sin{\frac{\pi}{x}})^n-1}{(1+\sin{\frac{\pi}{x}})^n+1}=1-\frac{2}{(1+\sin{\frac{\pi}{x}})^n+1}$
I thought that if $x\in Q$ and $x\in Q^c$, then we have to find discontinuity. I really need it to solve. Please help.
 A: If $\sin \frac \pi x > 0$ then $(1 + \sin \frac \pi x)^n \to \infty$. We know that $\sin \frac \pi x > 0$ if and only if $\frac \pi x \in (2k\pi, (2k+1)\pi)$ for every $k \in \Bbb Z$, i.e. $x \in (\frac 1 {2k+1}, \frac 1 {2k})$ for all $k \in \Bbb Z \setminus \{0\}$ or $x \in (1, \infty)$ (that comes from $k=0$). Since $x \in (0,1)$ we are left with only $x \in \bigcup _{k \ge 1} (\frac 1 {2k+1}, \frac 1 {2k})$.
If $\sin \frac \pi x = 0$ then $(1 + \sin \frac \pi x)^n =1$. This happens for $\frac \pi x = k\pi$ with $k \in \Bbb Z$, i.e. $x = \frac 1 k$ with $k \in \Bbb Z \setminus \{0\}$. Again, since $x \in (0,1)$ we keep only $x \in \{ \frac 1 k \mid k \in \Bbb N \setminus\{0\} \}$.
Finally, if $\sin \frac \pi x < 0$ then $(1 + \sin \frac \pi x)^n \to 0$. This happens for $\frac \pi x \in ((2k-1)\pi, 2k\pi)$ for all $k \in \Bbb Z$, i.e. for $x \in (\frac 1 {2k}, \frac 1 {2k-1})$ for all $k \in \Bbb Z \setminus \{0\}$ (the value $k=0$ produces negative numbers). Again, since $x \in (0,1)$ we keep only $x \in \bigcup _{k \ge 1} (\frac 1 {2k} \frac 1 {2k-1})$.
We deduce then that
$$\lim _{n \to \infty} \frac {(1 + \sin \frac \pi x)^n - 1} {(1 + \sin \frac \pi x)^n + 1} =
\begin{cases} 1, & x \in \bigcup _{k \ge 1} (\frac 1 {2k+1}, \frac 1 {2k}) \\
0, & x \in \{ \frac 1 k \mid k \in \Bbb N \setminus\{0\} \} \\
-1, & x \in \bigcup _{k \ge 1} (\frac 1 {2k}, \frac 1 {2k-1})
\end{cases}$$
and it is easy to see now that the points of discontinuity are the points $x = \frac 1 k$ with $k \in \Bbb N \setminus \{0\}$.
