Calculating $\lim_{x \rightarrow α}\frac{1 - \cos(ax^2+bx+c)}{(x-α)^2}$ If $α$ and $β$ are the roots of equation 
$$ax^2+bx+c = 0,$$ 
then find the following limit
$$\lim_{x \rightarrow α}\frac{1 - \cos(ax^2+bx+c)}{(x-α)^2}.$$
 A: Hint: There are two cases, $\alpha\ne \beta$ and $\alpha=\beta$. We can use L'Hospital's Rule twice, or expand the cosine in a power series around $x=\alpha$.  
Alternately, we can use various trigonometric identities. For example, one could use the fact that $1-\cos 2t=2\sin^2 t$. That way, we only need to use L'Hospital's Rule once. 
It may be useful to note that $ax^2+bx+c=a(x-\alpha)(x-\beta)$.
A: Hints:
1) Note that $ax^2+bx+c=a(x-\alpha)(x-\beta)$
2) As I wrote in my comment, substitute $t=x-\alpha$.
3) $$\lim_{z\to 0}\frac{1-\cos z}{z^2}=\frac12$$ which $z$ should you choose?
Edit: Here is the complete solution
$$\lim_{x\to\alpha}\frac{1 - \cos(ax^2+bx+c)}{(x-\alpha)^2}=\lim_{x\to\alpha}\frac{1 - \cos(a(x-\alpha)(x-\beta))}{(x-\alpha)^2}=\lim_{t\to0}\frac{1 - \cos(at(t+\alpha-\beta))}{t^2}$$
Denote $\gamma=\beta-\alpha$
$$\begin{align*}L&=\lim_{t\to0}\frac{1 - \cos(at(t-\gamma))}{t^2}\frac{a^2(t-\gamma)^2}{a^2(t-\gamma)^2}=\lim_{t\to0}\frac{1 - \cos(at(t-\gamma))}{(at(t-\gamma))^2}\lim_{t\to0}a^2(t-\gamma)^2\\
&=\lim_{z\to0}\frac{1 - \cos z}{z^2}\lim_{t\to0}a^2(t-\gamma)^2=\frac{a^2(\beta-\alpha)^2}{2}
\end{align*}$$
A: Hint: 1) Since you know the roots of the equation there maybe is a different way to write the equation (factorisation).
2) For the limit: Try L'Hôpital
A: $\frac12(\alpha-\beta)^2$
Start by  putting $t=x-\alpha$, and then solve. 
