Limit of an interception point as one parameter changes.

Question

Consider the functions \begin{align*} h_1(z)&=m\bigg(\frac{1}{z^3} + \frac{2e}{z^4}\bigg)\\ h_2(z)&=\frac{-M}{(a^2+z^2)^{3/2}} \end{align*} where $M$, $m$ and $e$ are real parameters and $e<0$. I know that these two only have one interception point, $z_0$ (I checked it graphically). I also know that the limit of $z_0$ as $m$ approaches infinity is equals to $2e$ (It was also checked graphically). BUT, if possible, I'd like to show it analytically. i.e, show that $$\lim_{m \rightarrow \infty} z_0 = 2e$$ Is it possible to show it by a "non numerical" approach? Any tips/help will be very appreciated.

Answer 1

It should actually be $-2e$, which assuming that $e<0$ will be positive. We can show this by considering solutions to $h_1(z) = h_2(z)$, since a point $z = z_0$ which satisfies this equation is an intercept. So, we have \begin{align*} h_1(z) &= h_{2}(z)\\[5pt] \implies m\bigg(\frac{1}{z^3} + \frac{2e}{z^4}\bigg)&=\frac{-M}{(a^2+z^2)^{3/2}}\\[5pt] \implies \frac{1}{z^3} + \frac{2e}{z^4} &= \frac{-M/m}{(a^2+z^2)^{3/2}}. \end{align*}

Now, taking the limit of each side as $m\to \infty$: $$\lim_{m\to\infty}\bigg(\frac{1}{z^3} + \frac{2e}{z^4}\bigg) =\lim_{m\to\infty}\left(\frac{-M/m}{(a^2+z^2)^{3/2}}\right) = 0. $$

So, as $m\to\infty$, we have that $z_0$ solves \begin{align*} \frac{1}{z_0^3} + \frac{2e}{z_0^4} &= 0\\ \implies\frac{1}{z_0^3} &= -\frac{2e}{z_0^4}\\ \implies z_0 &= -2e. \end{align*}