Prove $n \rightarrow 2^+$ as $a \rightarrow 0$

Question

While studying solids of revolution at college, I came across a problem related to physics that seems to have an answer difficult to prove mathematically, which I have not been able to obtain.

Motivation

Consider the surface $z=|x|^n+|y|^n$ for $0 \le z \le h$ and $n \ge 1$. If this surface were a physical object standing in a table, it is very clear that for values of $n \approx 1$ it will be in an unstable equilibrium, while at great values, such as $n=10$ it will be at an stable equilibrium, that is, the object may or may not tumble/fall under any force $F>0$ depending on the value of $n$. The question is then at which value of $n$ does it transition from unstable to stable?

$n=1.5$ and $n=10$

Problem statement

From the motivating problem I was able to obtain the equation below. Solving for $n$ as $a \rightarrow 0$ gives the answer, for any constant $c>0$

in the physical problem, $c$ is the z-coordinate of the center of mass of the object and $a$ is associated with at which point it will tumble. Taking $a \rightarrow 0$ provides a situation in which it may fall if moved by any infinitesimal amount, which is an unstable equilibrium

$$\left (c-a^n \right )na^{n-2}=1$$ However, there seems to be no feasible way to isolate $n$ in order to apply the limit $a \rightarrow 0$. Using Wolfram Alpha, it seems the value of $n$ approaches $2$ by using $a \approx 0$, but I see no way to prove it.

Answer 1

To avoid problems with noninteger powers of negative numbers, we assume $a>0$. Let $\varepsilon\in (0,1]$ be any number. If $n\ge 2+\varepsilon$, $a<\min\left\{e^{-1/2}, (c(2+\varepsilon))^{-1/\varepsilon}\right\}$ then $\left (c-a^n \right )na^{n-2}<cna^{n-2}\le c(2+\varepsilon)a^{\varepsilon}\le 1.$ The previous to the last inequality holds because the function $cna^{n-2}$ has the derivative $ca^{n-2}(1+n\ln a)$, and so is decreasing when $a$ is a fixed numbers less then $e^{-1/2}$. On the other hand, if $n\le 2-\varepsilon$, $a<\min\left\{1, \left(\frac c2\right)^{1/\varepsilon}\right\}$ then $\left (c-a^n \right )na^{n-2}\ge \frac{c}{2}a^{-\varepsilon}>1$. It follows $n\to 2$ when $a\to 0$.