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.
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.
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.
Expanding Lief's answer.
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. $