Questions mathématiques diverses

Question 94:
Soient $\alpha$ et $\beta$ deux solutions de l'équation $ax^2+bx+c=0$.
Déterminer alors la limite suivante :
$\lim\limits_{x \to \alpha}\dfrac{1-\cos\left({ax^2+bx+c}\right)}{(x-\alpha)^2}$.

Réponse 94:
$$\begin{align*} &\lim\limits_{x \to \alpha}\dfrac{1-\cos\left({ax^2+bx+c}\right)}{(x-\alpha)^2}\\ =&\lim\limits_{x \to \alpha}\dfrac{2\sin^2\left({\dfrac{ax^2+bx+c}{2}}\right)}{(x-\alpha)^2}\\ =&\lim\limits_{x \to \alpha}\dfrac{2\sin^2\left({\dfrac{a(x-\alpha)(x-\beta)}{2}}\right)}{(x-\alpha)^2}\\ =&\lim\limits_{x \to \alpha}2\left({\dfrac{\sin\left({\dfrac{a(x-\alpha)(x-\beta)}{2}}\right)}{\dfrac{a(x-\alpha)(x-\beta)}{2}}}\right)^2\dfrac{\left({\dfrac{a(x-\alpha)(x-\beta)}{2}}\right)^2}{(x-\alpha)^2}\\ =&\lim\limits_{X \to 0}\left({\dfrac{\sin X}{X}}\right)^2\times\lim\limits_{x \to \alpha} \dfrac{2a^2(x-\alpha)^2(x-\beta)^2}{4(x-\alpha)^2}\\ =&\lim\limits_{X \to 0}\dfrac{a^2(x-\beta)^2}{2}=\boxed{\dfrac{a^2(\alpha-\beta)^2}{2}} \end{align*}$$