# How, $f(x)=1/[1 + e^{1/\sin({n!{\pi}x})}]$ can be made discontinuous at any rational point in$[0,1]$?

How can I prove that the function $f$ defined by

$$f(x)=1/[1 + e^{1/\sin({n!{\pi}x})}]$$ Can be made discontinuous at any rational point in$[0,1]$ by a proper choice of $n$.

Plz help me with this.

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How $f(x)$ depends on $n$? Or $\lim\limits_{n\rightarrow{\infty}}$ is omitted? –  M. Strochyk Oct 27 '12 at 16:26

By noting that $f(x)$ is discontinuous whenever argument to sin is near zero (or multiples of $\pi$) because if its $0^-$, exponent is over -inf and when it is $0^+$ exponent is over +inf. Now if x were any rational, you could always choose any n to make the argument some multiple of $\pi$ and hence the function discontinuous.
For example, for $a=1/3$ choose $n=3$ so that $n!a$ is an integer. Then $$\lim_{x\to a^+}\frac{1}{1+\exp(1/\sin(3!\pi x))} = \frac{1}{1+\infty} = 0$$ but $$\lim_{x\to a^-}\frac{1}{1+\exp(1/\sin(3!\pi x))} = \frac{1}{1+0} = 1.$$