<s>[INST] Limit of an interception point as one parameter changes. 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.
 [/INST] 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*}
</s>