Asymptotic for the roots of a Polynomial

I am studying an eigenvalue problem on an Hilbert space. I turn it into a first order dynamical system. I need the asymptotic behavior of that asymptotic system. To do that, I am faced with computing the eigenvalues associated to that asymptotic system. For some specific values of my parameters, the characteristic polynomial takes the form of the following polynomial for the variable $r$ $$ (1.6)r^3-\lambda r^2-(0.4) r+\lambda=0, $$ where $\lambda$ is a complex parameter. I know that, as $|\lambda|\to \infty$, $\Re(\lambda)>0$, then the real parts of the roots tend to $\pm 1$ and $0$. I know this by numerics. Usually, when faced with such a problem, I can deal with it through singular perturbations assuming $\lambda=\lambda_0/\epsilon$, but that case seems more difficult to handle than other cases I worked on. If someone could help. Even proving this fact for $\lambda$ real would be helpful.

Fact 1. Let $0 \ne \lambda \in \mathbb{C}$. Let $r_1, r_2, r_3$ with $|r_1| \le |r_2| \le |r_3|$ and $\mathrm{Re}(r_1) \le \mathrm{Re}(r_2)$ be the roots of the cubic equation $$\frac85 r^3 - \lambda r^2 - \frac25 r + \lambda = 0.$$ Then, $$\lim_{|\lambda| \to \infty} r_1 = -1, \quad \lim_{|\lambda| \to \infty} r_2 = 1, \quad \lim_{|\lambda| \to \infty} \frac{r_3}{\lambda} = \frac58.$$

Proof of Fact 1.

By Vieta's theorem, we have \begin{align*} r_1 + r_2 + r_3 &= \frac58 \lambda, \tag{1}\\ r_1r_2 + r_2r_3 + r_3 r_1 &= - \frac14,\tag{2}\\ r_1r_2r_3 &= - \frac58 \lambda.\tag{3} \end{align*}

We claim that $|r_2| < 2$. Indeed, from (1) and (3), we have $$r_1 + r_2 + r_3 + r_1 r_2 r_3 = 0. \tag{4}$$ $\frac{4(1 + r_2r_3)}{r_2^2r_3^2} \times (2) - \frac{4(r_2 + r_3)}{r_2^2r_3^2}\times (4)$ yields $$4 = \frac{4}{r_2^2} + \frac{4}{r_3^2} + \frac{3}{r_2r_3} - \frac{1}{r_2^2r_3^2}.$$ However, $|\frac{4}{r_2^2} + \frac{4}{r_3^2} + \frac{3}{r_2r_3} - \frac{1}{r_2^2r_3^2}| \le \frac{4}{|r_2|^2} + \frac{4}{|r_3|^2} + \frac{3}{|r_2|\cdot |r_3|} + \frac{1}{|r_2|^2|r_3|^2} < 3$ if $2 \le |r_2| \le |r_3|$. The claim is proved.

From (1), we have $$\frac{r_1 + r_2}{\lambda} + \frac{r_3}{\lambda} = \frac58$$ which results in (using $|r_1|\le |r_2| < 2$) $$\lim_{|\lambda| \to \infty} \frac{r_3}{\lambda} = \frac58. \tag{5}$$

From (3), we have $$r_1r_2 = -\frac58\cdot \frac{\lambda}{r_3}$$ which results in (using (5)) $$\lim_{|\lambda| \to \infty} r_1r_2 = -1. \tag{6}$$

From (2), we have $$r_1 + r_2 = - \frac{1}{4r_3} - \frac{r_1r_2}{r_3}$$ which results in (using (5)) $$\lim_{|\lambda| \to \infty} (r_1 + r_2) = 0. \tag{7}$$

From (6) and (7), we have $$\lim_{|\lambda| \to \infty} r_1 = -1, \quad \lim_{|\lambda| \to \infty} r_2 = 1.$$

We are done.

For real $\lambda$, the discriminant of the cubic equation is $$\Delta=\frac{4}{625} \left(625 \lambda ^4-8975 \lambda^2+64\right)$$ which is positive as soon as $\lambda>4$; so, for large $\lambda$, three real roots.

Using the trigonometric solution, the roots are, for $k=0,1,2$, $$r_k=\frac{1}{24} \left(5 \lambda +2 \sqrt{25 \lambda ^2+48} \cos \left(\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(\frac{5 \lambda \left(25 \lambda ^2-792\right)}{\left(25 \lambda ^2+48\right)^{3/2}}\right)\right)\right)\right)$$ Expanded as series, the asymptotics are

$$r_0=\frac{5 \lambda }{8}-\frac{6}{5 \lambda }+O\left(\frac{1}{\lambda^3}\right)$$ $$r_1=1+\frac{3}{5 \lambda }+\frac{57}{50 \lambda ^2}+O\left(\frac{1}{\lambda^3}\right)$$ $$r_2=-1+\frac{3}{5 \lambda }-\frac{57}{50 \lambda ^2}+O\left(\frac{1}{\lambda^3}\right)$$

Edit

If $\lambda=(a+i\,b)$, expanding for large value of $a$ $$r_0=\frac{5}{8} (a+i b)-\frac{6}{5 a}+\frac{6 i b}{5 a^2}+\frac{6 \left(25 b^2-112\right)}{125 a^3}+O\left(\frac{1}{a^4}\right)$$ $$r_1=1+\frac{3}{5 a}+\frac{3 (19-10 i b)}{50 a^2}-\frac{3 \left(25 b^2+95 i b-112\right)}{125 a^3}+O\left(\frac{1}{a^4}\right)$$ $$r_2=-1+\frac{3}{5 a}-\frac{3(19+10ib)}{50 a^2}-\frac{3 \left(25 b^2-95 i b-112\right)}{125 a^3}+O\left(\frac{1}{a^4}\right)$$

Update

In comments @RiverLi smartly proposed to use Newton method to generate asymptotics of the roots.

Using the idea with higher order methods, we have $$r_0=\frac{5 \lambda }{8}-\frac{6}{5 \lambda}-\frac{672}{125 \lambda ^3}-\frac{93184}{3125 \lambda ^5}+O\left(\frac{1}{\lambda^7}\right)$$ $$r_1=1+\frac{3}{5 \lambda }+\frac{57}{50\lambda ^2}+\frac{336}{125 \lambda ^3}+\frac{7107}{1000 \lambda ^4}+O\left(\frac{1}{\lambda^5}\right)$$ $$r_2=-1+\frac{3}{5 \lambda }-\frac{57}{50\lambda ^2}+\frac{336}{125 \lambda^3}-\frac{7107}{1000 \lambda^4}+O\left(\frac{1}{\lambda^5}\right)$$