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Question Number 110736 by abdullahquwatan last updated on 30/Aug/20

given f(x)=ax^2 +bx+c and f(x)  is negative on a<x<b. The value of  lim_(x→a)  (((x−a)^2 )/(1−cos (f(x))))

$$\mathrm{given}\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}\:\mathrm{and}\:{f}\left({x}\right) \\ $$ $$\mathrm{is}\:\mathrm{negative}\:\mathrm{on}\:{a}<{x}<{b}.\:\mathrm{The}\:\mathrm{value}\:\mathrm{of} \\ $$ $$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{\left({x}−{a}\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{cos}\:\left({f}\left({x}\right)\right)} \\ $$

Commented byabdullahquwatan last updated on 30/Aug/20

answer in a and b

$$\mathrm{answer}\:\mathrm{in}\:{a}\:\mathrm{and}\:{b} \\ $$

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