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Question Number 122834 by liberty last updated on 20/Nov/20

If ax^2 +bx+c = 0 has the roots are   p and q then the value of lim_(x→p)  ((1−cos (ax^2 +bx+c))/((x−p)^2 ))   is ___

$${If}\:{ax}^{\mathrm{2}} +{bx}+{c}\:=\:\mathrm{0}\:{has}\:{the}\:{roots}\:{are}\: \\ $$$${p}\:{and}\:{q}\:{then}\:{the}\:{value}\:{of}\:\underset{{x}\rightarrow{p}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left({ax}^{\mathrm{2}} +{bx}+{c}\right)}{\left({x}−{p}\right)^{\mathrm{2}} }\: \\ $$$${is}\:\_\_\_ \\ $$

Commented by bemath last updated on 20/Nov/20

 lim_(x−p→0)  ((1−cos (ax^2 +bx+c))/((x−p)^2 )) =   lim_(x−p→0) ((2sin^2 (((ax^2 +bx+c)/2)))/((x−p)^2 )) =    2 × lim_(x−p→0)  (((sin (((a(x−p)(x−q))/2)))/(x−p)))^2 =   2 × (1/4)a^2 (p−q)^2  = ((a^2 (p−q)^2 )/2)

$$\:\underset{{x}−{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left({ax}^{\mathrm{2}} +{bx}+{c}\right)}{\left({x}−{p}\right)^{\mathrm{2}} }\:= \\ $$$$\:\underset{{x}−{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{{ax}^{\mathrm{2}} +{bx}+{c}}{\mathrm{2}}\right)}{\left({x}−{p}\right)^{\mathrm{2}} }\:=\: \\ $$$$\:\mathrm{2}\:×\:\underset{{x}−{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:\left(\frac{{a}\left({x}−{p}\right)\left({x}−{q}\right)}{\mathrm{2}}\right)}{{x}−{p}}\right)^{\mathrm{2}} = \\ $$$$\:\mathrm{2}\:×\:\frac{\mathrm{1}}{\mathrm{4}}{a}^{\mathrm{2}} \left({p}−{q}\right)^{\mathrm{2}} \:=\:\frac{{a}^{\mathrm{2}} \left({p}−{q}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$

Answered by $@y@m last updated on 20/Nov/20

(((p−q)^2 )/2)

$$\frac{\left({p}−{q}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$

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