if$a$ is a repeated root of  $a{x}^2+bx+c=0$, $\underset{x\to \alpha }{lim}\frac{\sin \left(a{x}^2+bx+c\right)}{(x-\alpha {)}^2}$ is equal to 

Since $a$ is a repeated root of $a{x}^2+bx+c=0$

$\therefore a{x}^2+bx+c=a(x-\alpha {)}^2$

$\begin{array}{l}\underset{x\to \alpha }{lim} \frac{\sin \left(a{x}^2+bx+c\right)}{(x-\alpha {)}^2}\\ =\underset{x\to \alpha }{lim} \frac{\sin a(x-\alpha {)}^2}{(x-\alpha {)}^2}=a\underset{x\to \alpha }{lim} \frac{\sin a(x-\alpha {)}^2}{a(x-\alpha {)}^2}=a\times 1=a.\end{array}$