Some calculations with polynomials in the proof of the Routh-Hurwitz test In an article on the Routh-Hurwitz test, I couldn't see why the following result is true:

Let $$p(s)=a_ns^n+a_{n-1}s^{n-1}+\dots+a_{1}s+a_0$$ and $$ \eta_{*} = \frac{a_n}{a_{n-1}}$$ $$g_{\eta}(s):=p(s)-\eta (a_{n-1}s^n+a_{n-3}s^{n-2}+a_{n-5}s^{n-4}+\cdots)$$  Then one and only one zero of $g_{\eta}(s)$ tends to $\frac{-a_{n-1}}{a_n-\eta a_{n-1}}$ as $\eta \in \mathbb{R}$ approaches $\eta_{*}$ from the origin.

Notice that when $\eta=0$ then $g_{\eta}(s)=p(s)$ and when $\eta=\frac{a_n}{a_{n-1}}$ then $$g(s)=p(s)-\frac{a_n}{a_{n-1}}(a_{n-1}s^n+a_{n-3}s^{n-2}+a_{n-5}s^{n-4}+\cdots)$$ a polynomial of degree $n-1$.
The title of the article is "Elementary proof of the Routh-Hurwitz test" (DOI link), and in the part quoted above, the author has already demonstrated that $p(s)$ and $g(s)$ have the same number of purely imaginary roots. I'd appreciate any help. Thanks in advance!
 A: Just to check: you do realize that
$$ \lim_{\eta\to \eta^*}\left|\frac{-a_{n-1}}{a_n - \eta a_{n-1}}\right| = \infty, $$
right? So what the author wants to say is that there is one zero that goes to $\infty$ asymptotically as this fraction does.
The polynomials $p(s)$ and $g_\eta(s), 0\leq \eta < \eta^*$ all have degree $n$ while $g_{\eta^*}(s)$ has degree $n-1$. Thus by general continuity arguments one of the roots has to wander off to $\infty$ as $\eta \to \eta^*$ - the question is, whether there is a reasonable asymptote. And yes, there is. For a general result on the asymptotes of the roots going to infinity in the case of degree differences, see Proposition 4.1.3 in Hinrichsen/Pritchard. Mathematical Systems Theory I, Springer.
In the present case, the situation is a bit easier and the original author actually makes a hint at the argument. We already know that it is exactly one root going to $\infty$. (This is what the author means when he writes "one and only one root" behaves like ...). And we can write
$$ g_\eta(s) = \left( a_n - \eta a_{n-1}\right)s^n + a_{n-1} s^{n-1} + ...$$
Then we could write
$$ g_\eta(s) = \left( (a_n - \eta a_{n-1})s + a_{n-1}\right) s^{n-1} + r_\eta(s),$$
where $n-2 \geq \mathrm{deg} r_\eta(s) =: r$.
Now introduce new variables
$\zeta = \gamma(\eta) s := \frac{-a_n + \eta a_{n-1}}{a_{n-1}}  s$.
Then
$$ g_\eta(s) = g_\eta(\gamma^{-1}(\eta) \zeta) = \left( - a_{n-1}\zeta + a_{n-1}\right) \gamma^{-(n-1)}(\eta)\zeta^{n-1} + r_\eta(\gamma^{-1}(\eta) \zeta) .$$
Now take out $\gamma^{-(n-1)}$. As $\mathrm{\deg}\ r_\eta \leq n-2$ this means that in every term of that polynomial a factor $\gamma(\eta)$ appears. Omitting the details, let's just write $r_\eta(\gamma^{-1} \zeta) = \gamma^{-(n-1)} \hat{r}_\eta(\gamma, \zeta)$ and note that the coefficients of $\hat{r}_\eta$ go to $0$ as $\eta\to\eta^*$.
$$ = \gamma^{-(n-1)}(\eta)\left[\left( - a_{n-1}\zeta + a_{n-1}\right) \zeta^{n-1} + \hat{r}_\eta(\gamma(\eta), \zeta) \right].$$
Now we are done. As $\eta\to\eta^*$, we have $\hat{r} \to 0$, $\gamma^{-(n-1)}$ does not influence the zeros  and by continuity the zeros of the polynomial in $\zeta$ approach $1$ and $n-1$ zeros.
Transforming back, this means that there is a branch $s(\eta)$ of the zero set of $g_\eta(s)$ such that
$$ \lim_{\eta\to\eta^*} 1 - \frac{-a_n + \eta a_{n-1}}{a_{n-1}}  s(\eta) = 0 .$$
This is what was meant, and all that is required for that particular proof.
