Let $\alpha$. and $\beta$ be the roots of $a{x}^2+bx+c=0$, then $\underset{x\to a}{lim} \frac{1-\cos \left(a{x}^2+bx+c\right)}{(x-\alpha {)}^2}$ is equal to

Solution:

It is given that ${\alpha }_{,}\beta$ are roots of $a{x}^2+bx+c$

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

Now, $\underset{x\to \alpha }{lim} \frac{1-\cos \left(a{x}^2+bx+c\right)}{(x-\alpha {)}^2}$

$\begin{array}{l}=2\underset{x\to \alpha }{lim} \frac{{\sin }^2\left\{\frac{\left(a{x}^2+bx+c\right)}{2}\right\}}{(x-\alpha {)}^2}\\ \left.=2\underset{x\to \alpha }{lim} \frac{{\sin }^2\left\{\frac{a(x-\alpha )(x-\beta )}{2}\right\}}{(x-\alpha {)}^2}\right\}\\ =2\underset{x\to \alpha }{lim} \left[\frac{{\left.\sin \left\{\frac{a(x-\alpha )(x-\beta )}{2}\right\}\right]}^2}{\frac{a(x-\alpha )(x-\beta )}{2}}\right]\times \frac{a^2}{4}(x-\beta {)}^2\\ =2(1{)}^2\times \frac{a^2}{4}(\alpha -\beta {)}^2=\frac{a^2}{2}(\alpha -\beta {)}^2\end{array}$