106x2+bx+c = (x-α)(x-β)
=> lim(x→β) 2sin2[(x-α)(x-β)/2]/(x-β)2
= lim(x→β)2(x-α)2/4
= (β-α)2/2
If α, β are the roots of the equation x2 + bx + c = 0 then find
lim (1 - cos(x2 + bx + c))/(β - x)2
x->β
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2 Answers
106
