show that $f(x) = \frac{1}{1+e^{\frac{1}{n!\pi x}}}$ can be made discontinuous at any rational point in interval (0,1) by proper selection of n

Question

show that $f(x) = \frac{1}{1+e^{\frac{1}{n!\pi x}}}$ can be made discontinuous at any rational point in interval (0,1) by proper selection of n

can you tell me is my method correct?

Take 2 sequences $x_n=\frac{1}{n!} < \frac{1}{2^{(n-1)}} \to 0, y_n=\frac{1}{(n-1)!} \to 0$

Now $f(x_n) = \frac{1}{1+e^{\frac{1}{\pi}}} = 0.42.., f(y_n)=\frac{1}{1+e^{\frac{1}{(n-1)\pi}}}= 1/2 \implies f(x_n) \neq f(y_n)$. So f is not continuous at x=0+

Pls correct me if i am wrong ...

Answer 1

As mentioned by Hagen in the comments, $f$ is continuous on $(0,1)$ for every $n$.

If $g$ and $h$ are continuous in I and $h\neq 0$ then $g/h$ is continuous in I; so $f(x)=\frac{1}{1+e^{\frac{1}{\pi x n!}}}$ is continuos in $(0,1)$.

I think you maybe could have do a little confusion with the fact that the $n$ that appears in $f$ is actually a $\textbf{fixed parameter}$, while the $n$ of the indexes of the successions are supposed to change.

Just change the name of the parameter $n$ in $f$, let's say we define $f(x)=\frac{1}{1+e^{\frac{1}{\pi x K!}}}$ with $K \in \mathbb{N}$ $\textbf{fixed}$, and think again to your example.