I am given a quadric equation such that $ax^2 + bx +c=0$ whose roots are $\alpha$ and $\beta$ then what would be value $$\lim\limits_{x \to \alpha} \frac{1-\cos( ax^2 + bx +c) }{(x-\alpha)^2}$$ Now since $x$ is tending to root of input in $\cos$ so my limits become $0/0$ form so I applied L'Hospital Rule hence my limit becomes $$\lim\limits_{x \to \alpha} \frac{\sin( ax^2 + bx +c).2ax +b }{2(x-\alpha)}$$ now since $\alpha$ and $\beta$ are roots hence my expression can be written as $$\lim\limits_{x \to \alpha} \frac{ (x-\beta ) \sin( (x-\alpha)(x-\beta))(2ax +b)}{2(x-\alpha) (x-\beta ) }$$ now it becomes of form $\frac{\sin x}{x}$ when $x$ approaches $0$ so finally I reach $$\lim\limits_{x \to \alpha} \frac{(2ax +b)(x-\beta )}{2}$$ which finally becomes $$ \frac{(2a\alpha +b)(\alpha-\beta )}{2}$$ But my answer does not matches , what did I do wrong?
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The mistake is
$$ax^2+bx+c=a(x-\alpha)(x-\beta)$$
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Or faster : $ax^2+bx+c=a(x−α)(x−β)$
1-cos(u) ~ $\frac{u^2}{2}$ , when u->0
=> your expression is equivalent to : $\frac{a^2*(x-α)^2(x-β)^2}{2*(x-α)^2}$ when x-> α
So the final equivalent is : $\frac{a^2*(x-β)^2}{2}$ which is (I hope) the answer
mvggz
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Thx ;) , I had to do my part in this kind of calculus when I was younger – mvggz Oct 15 '14 at 14:40