Question:
How to intuitively see that $\frac{x^n}{K^n + x^n}$ is a sigmoid for large n, where $0<K<1$?
For example, I have chosen $K=0.5$ and plotted the graph of $f(x) = \frac{x^n}{K^n + x^n}$ for different $n$s, and I have learned that when $n\to \infty$, $f\to H$, heavyside step function. But, I cannot understand why does this function behave differently for $x< K$ and $K<x$.
I will write $f_{n}(x) = \dfrac{x^{n}}{K^{n} + x^{n}}.$ Note that $f_{n}(0) = 0$ and $f_{n}(K) = \dfrac{1}{2}$ for all $n \geq 1$, and for $x \neq 0$ we can divide by $x^{n}$ to get \begin{equation*} f_{n}(x) = \frac{1}{1 + \mathopen{}\left(\frac{K}{x}\right)^{n}\mathclose{}}. \end{equation*} So for $\dfrac{K}{x} < 1$, \begin{equation*} f_{n}(x) \to \frac{1}{1 + 0} \end{equation*} as $n \to \infty$. For $\dfrac{K}{x} > 1$, \begin{equation*} f_{n}(x) \to 0 \end{equation*} as $n \to \infty$ due to the denominator becoming large.
Note also that if $n$ is odd, $f_{n}$ will have a pole at $x = -K.$