If $\alpha \ne 0$ is a real root of the cubic equation $a{x}^3+b{x}^2+bx+a=0$ then the value of $\underset{x\to \frac{1}{\alpha }}{\lim }\frac{\tan \left(a{x}^3+b{x}^2+bx+a\right)}{(\alpha x-1)}$ is where $a\ne 0,b\in R$

If $\alpha \ne 0$ is a real root of the  reciprocal equation $a{x}^3+b{x}^2+bx+a=0 then \frac{1}{\alpha }$ is also a root

$\text{ Clearly }x=-1\text{ is a root since f(-1)=0}$

$\underset{x\to \frac{1}{\alpha }}{\lim }\frac{\tan \left(a{x}^3+b{x}^2+bx+a\right)}{(\alpha x-1)}$

$$\begin{gathered} =\underset{x\to \frac{1}{\alpha }}{\lim }\frac{\tan \left(a(x+1)(x-\alpha )\left(x-\frac{1}{\alpha }\right)\right)}{(\alpha x-1)} \\ =\underset{x\to \frac{1}{\alpha }}{\lim }\frac{\tan \left(a(x+1)(x-\alpha )\left(x-\frac{1}{\alpha }\right)\right)}{a(x+1)(x-\alpha )(x-{\displaystyle \frac{1}{\alpha }})}\left(\frac{a(x+1)(x-\alpha )}{\alpha }\right) \\ =\frac{a}{\alpha }(\frac{1}{\alpha }+1)(\frac{1}{\alpha }-\alpha )=\frac{a\left(1+\alpha \right)\left(1-{\alpha }^2\right)}{{\alpha }^3}=\frac{a\left(1+\alpha \right)\left(1-\alpha \right)\left(1+\alpha \right)}{{\alpha }^3}=\frac{a{\left(1+\alpha \right)}^2\left(1-\alpha \right)}{{\alpha }^3} \end{gathered}$$