Solutions to Cubic equation that dont diverge when reduced to linear equation

[phil ess]

I am currently trying to solve for the metric function for a black hole in adS space with quasi-topological gravity. The details aren't too important, but the point is that I have to solve for a cubic at one point, and choose the correct solution, which is the one that reduces to a linear equation when the cubic and squared coefficients are set to zero.

Consider solving the quadratic case as an example of what I'm trying to say:

The solution to $ax^{2}+b^{x}+c=0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

In this case I would be interested in the solution that doesn't diverge when $a\rightarrow0$ which reduces the quadratic to a linear equation. In this case we would take the positive of the two solutions, because then as $a\rightarrow0$ we would have $\sqrt{b^{2}-4ac}\rightarrow b$ and thus $x\rightarrow\frac{0}{0}$ and does not diverge as the negative solution would.

I am trying to do the analogous thing for a cubic equation, where I have to decide which of the three solutions will give this type of behaviour as the coefficients of the cubic and quadratic terms are taken to zero like in the above example.

Any help would be greatly appreciated. Thanks!

 

[mathman]

I am currently trying to solve for the metric function for a black hole in adS space with quasi-topological gravity. The details aren't too important, but the point is that I have to solve for a cubic at one point, and choose the correct solution, which is the one that reduces to a linear equation when the cubic and squared coefficients are set to zero.

Consider solving the quadratic case as an example of what I'm trying to say:

The solution to $ax^{2}+b^{x}+c=0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

In this case I would be interested in the solution that doesn't diverge when $a\rightarrow0$ which reduces the quadratic to a linear equation. In this case we would take the positive of the two solutions, because then as $a\rightarrow0$ we would have $\sqrt{b^{2}-4ac}\rightarrow b$ and thus $x\rightarrow\frac{0}{0}$ and does not diverge as the negative solution would.

I am trying to do the analogous thing for a cubic equation, where I have to decide which of the three solutions will give this type of behaviour as the coefficients of the cubic and quadratic terms are taken to zero like in the above example.

Any help would be greatly appreciated. Thanks!

$x=\frac{2c}{-b\pm\sqrt{b^{2}-4ac}}$

Above is equivalent. When a -> 0, x -> -c/b or becomes infinite (two roots).