<s>[INST] Does the limit of the cubic formula approach the quadratic one as the cubic coefficient goes to $0$? The formula for solving a cubic equation of the form $ax^3+bx^2+cx+d=0$ does not seem to yield the quadratic formula for the limit $\lim _{a \rightarrow 0} \text{(cubic formula)}$.
But, if one tries the same thing with the quadratic formula the limit exists for the right choice of the square root sign.
My question is, is there a way to take the limit $\lim _{a \rightarrow 0} \text{(cubic formula)}$ and produce the quadratic formula? or does the limit simply not exist? Finally, if the limit does not exist, is there a technical reason for that?
Any input is very much appreciated. Thanks
Edit:
I was trying to tackle the simple case of $b=0$. Doing so, the cubic formula reduces to
$$\left(-\frac{d}{2a} - \left(\frac{4c^3 + 27ad^2}{108a^3}\right)^{\frac{1}{2}}\right)^{\frac{1}{3}} + \left(-\frac{d}{2a} + \left(\frac{4c^3 + 27ad^2}{108a^3}\right)^{\frac{1}{2}}\right)^{\frac{1}{3}}$$
If one only considers the first term cubed, then it can be written as
$$\frac{1}{a}\left(-\frac{d}{2} - \left(4c^3+27ad^2\right)^{\frac{1}{2}}\right)$$
and I don't see how is it possible to find a finite limit as $a\rightarrow 0$. Am I missing something trivial?
 [/INST] When $a \to 0$ one of the roots goes to infinity, which complicates the algebraic manipulations. Instead, it is easier to show that when $d=0$ the non-zero roots reduce to the quadratic formula.
Assume WLOG $\,a=1\,$, then with the wikipedia notations for the cubic formula in the case $d=0\,$:
$$
\begin{align}
\Delta_{0} = b^{2}-3ac &= b^2 - 3c
\\ \Delta_{1} = 2b^{3}-9abc+27a^{2}d &= 2b^3 - 9bc
\\ C = {\sqrt[{3}]{\frac{1}{2}\left(\Delta _{1}\pm {\sqrt {\Delta _{1}^{2}-4\Delta _{0}^{3}}}\right)}} &= \sqrt[3]{\frac{1}{2}\left(2b^3 - 9bc \pm 3c\sqrt{-3(b^2 - 4c)}\right)} \tag{1}
\end{align}
$$
The radicals in $(1)$ can be denested, since it is straightforward to verify that:
$$
\left(-b \pm \sqrt{-3(b^2-4c)}\right)^3 = 4\left(2b^3 - 9bc \pm 3c\sqrt{-3(b^2 - 4c)}\right) \tag{†}
$$
It follows that:
$$
\require{cancel}
\begin{align}
C &= \frac{1}{2}\left(-b \pm \sqrt{-3(b^2-4c)}\right) \tag{2}
\\ \frac{\Delta_0}{C} = \frac{2(b^2-3c)}{-b \pm \sqrt{-3(b^2-4c)}} &= \frac{1}{2}\left(-b \mp \sqrt{-3(b^2-4c)}\right) \tag{3}
\end{align}
$$
Finally, the roots are $\displaystyle\, x = -{\frac {1}{3}}\left(b + \omega C+{\frac {\Delta_0}{\omega C}}\right)\,$ where $\,\omega^3 = 1\,$, and $\,\omega = 1\,$ gives the root $\,x=0\,$. Otherwise, with $\,\omega = \dfrac{-1 \pm i\sqrt{3}}{2}\,$ a complex cube root of unity, using $(2)$ and $(3)\,$:
$$
\begin{align}
x &= -\frac{1}{3}\left(b + \omega\,\frac{-b \pm \sqrt{-3(b^2-4c)}}{2} + \overline\omega\,\frac{-b \mp \sqrt{-3(b^2-4c)}}{2}\right)
\\ &= -\frac{1}{3}\left(b \cdot \left(1-\frac{\omega + \overline\omega}{2}\right) \pm \sqrt{-3(b^2-4c)} \cdot \frac{\omega - \overline\omega}{2} \right)
\\ &= -\frac{1}{3}\left(\frac{3}{2}\, b \pm \frac{i \sqrt{3}}{2} \sqrt{-3(b^2-4c)} \right)
\\ &= \frac{-b \pm \sqrt{b^2-4c}}{2}
\end{align}
$$
The latter matches the quadratic formula, as expected.

[ EDIT ] $\;$ The denesting of $\,C = {\sqrt[{3}]{\frac{\Delta _{1}}{2}\pm {\sqrt {\frac{\Delta _{1}^{2}}{4} - \Delta _{0}^{3}}}}}\,$ in $\,(\dagger)\,$ follows from my answer here.

a sufficient condition for $\,\sqrt[3]{m \sqrt{p} \pm n\sqrt{q}}\,$ to denest is for $\,m^2 \cdot p - n^2 \cdot q\,$ to be the cube of a rational $\,r\,$, and for the cubic $\,p\, t'^{\,3} - 3r\, t' - 2m\,$ to have an appropriate rational root

Replacing "rational" with "rational expressions in the coefficients", the sufficient conditions are satisfied in this case, since:

*

*$\,m^2 \cdot p - n^2 \cdot q = \Delta_0^3\,$ is a perfect cube;


*$\,t'^{\,3} - 3 \Delta_0 t' - \Delta_1 = t'^{\,3} - 3(b^2-3c) t' - b(2b^2-9c)\,$ has the root $\,t'=-b\,$.
</s>