. Since α is a root of the quadratic equation ax2+bx+c=0, we know that aα2+bα+c=0. Therefore, near x=α, we can express ax2+bx+c as follows: ax2+bx+c=a(x−α)(x−β) where β is the other root. We are interested in the behavior of 1−cos(ax2+bx+c) as x approaches α. Let's rewrite ax2+bx+c near x=α : Since α is a root: ax2+bx+c≈a(x−α)(x−β) Thus, as x⟶α : ax2+bx+c⟶a(α−α)(α−β)=0 Let's apply the Taylor series expansion for cos(y) around y=0 :
cos(y)≈1−‌
y2
2
Replacing y by a(x−α)(x−β) in the cosine function: cos(a(x−α)(x−β))≈1−‌
(a(x−α)(x−β))2
2
Therefore, we have:
1−cos(ax2+bx+c)≈1−(1−‌
(a(x−α)(x−β))2
2
)=‌
(a(x−α)(x−β))2
2
Now, substituting this back into the limit expression: