Why is the purple function equal to $ex/\ln(x)?$ How do I prove this?
After tracing out successive blue functions I wanted to find the function that passed through the minimum points of the successive blue curves. After a while I found it. But I'm not sure why the constant factor is equal to $e$. Can someone shed light on this?
Let
$$f\left(x\right) = \cfrac{x}{\ln x} \tag{1}\label{eq1}$$ $$g\left(x\right) = \cfrac{N - x}{\ln\left(N - x\right)} \tag{2}\label{eq2}$$ $$h\left(x\right) = f\left(x\right)g\left(x\right) \tag{3}\label{eq3}$$
You want to know why the limiting behavior of $h\left(x\right)$ is $e f\left(x\right)$. To see this, check where the derivative of $g\left(x\right)$ is $0$. Using the product & chain rules, we get that
\begin{align} g'\left(x\right) & = -\cfrac{1}{\ln\left(N - x\right)} + \cfrac{\left(N - x\right)\left(-1\right)\left(-1\right)}{\ln^2 \left(N - x\right)\left(N - x\right)} \\ & = -\cfrac{\ln\left(N - x\right)}{\ln^2\left(N - x\right)} + \cfrac{1}{\ln^2\left(N - x\right)} \\ & = \cfrac{1 - \ln\left(N - x\right)}{\ln^2 \left(N - x\right)} \tag{4}\label{eq4} \end{align}
Assuming $\ln\left(N - x\right) \neq 0$, this gives that
$$\ln\left(N - x\right) = 1 \Rightarrow N - x = e \tag{5}\label{eq5}$$
Plugging this into \eqref{eq2} gives
$$g\left(e\right) = \cfrac{e}{\ln\left(e\right)} = e \tag{6}\label{eq6}$$
You could differentiate \eqref{eq4} again to show the $2$nd derivative is positive at $x = e$. Alternatively, since $\ln\left(N - x\right)$ goes to $0$ as $N - x$ goes to $1^+$, this indicates that $g\left(x\right)$ goes to $\infty$. Similarly, as $N - x$ goes to $\infty$, $g\left(x\right)$ also goes to $\infty$. This shows that \eqref{eq6} is a local minimum. As such, the equation
$$m\left(x\right) = \cfrac{ex}{\ln x} \tag{7}\label{eq7}$$
provides a lower bound for those sections of the various functions in \eqref{eq3}.