Question

If α is a repeated root of ax² + bx + c = 0, then $\underset{x\to \alpha }{\lim }\frac{\tan \left(a{x}^2+bx+c\right)}{{\left(x-\alpha \right)}^2}$ is
(a) a
(b) b
(c) c
(d) 0

Answer 1

(a)

$$\begin{gathered} \underset{x\to \alpha }{\lim }\left[\frac{\tan \left(a{x}^2+bx+c\right)}{{\left(x-\alpha \right)}^2}\right] \\ =\underset{x\to \alpha }{\lim }\left[\frac{\tan \left\{a\left(x-\alpha \right)\left(x-\alpha \right)\right\}}{a{\left(x-\alpha \right)}^2}\right]\times a \\ =\underset{x\to \alpha }{\lim }\left[\frac{\tan \left\{a{\left(x-\alpha \right)}^2\right\}}{a{\left(x-\alpha \right)}^2}\right]\times a \\ =1\times a\left[\underset{\theta \to 0}{\lim }\left(\frac{\tan \theta }{\theta }\right)=1\right] \\ =a \end{gathered}$$