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Question Number 63920 by raj last updated on 11/Jul/19
If α,β are root of quadratic equation ax^2 +bx+c then lim_(x→α) ((1−cos (ax^2 +bx+c))/((x−α)^2 ))=?
$$\mathrm{If}\:\alpha,\beta\:\mathrm{are}\:\mathrm{root}\:\mathrm{of}\:\mathrm{quadratic}\:\mathrm{equation} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}\:\mathrm{then} \\ $$$$\underset{{x}\rightarrow\alpha} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\left({ax}^{\mathrm{2}} +{bx}+{c}\right)}{\left({x}−\alpha\right)^{\mathrm{2}} }=? \\ $$
Commented by Prithwish sen last updated on 11/Jul/19
=Lt[((sin(((x−α)(x−β))/2))/(((x−α)(x−β))/2))]^2 .(((x−β)^2 )/2) =(((α−β)^2 )/2) please check.
$$=\mathrm{Lt}\left[\frac{\mathrm{sin}\frac{\left(\mathrm{x}−\alpha\right)\left(\mathrm{x}−\beta\right)}{\mathrm{2}}}{\frac{\left(\mathrm{x}−\alpha\right)\left(\mathrm{x}−\beta\right)}{\mathrm{2}}}\right]^{\mathrm{2}} .\frac{\left(\mathrm{x}−\beta\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{\left(\alpha−\beta\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{please}\:\mathrm{check}. \\ $$
Commented by kaivan.ahmadi last updated on 11/Jul/19
hop lim_(x→α) (((2ax+b)sin(ax^2 +bx+c))/(2(x−α)))=^(hop) lim_(x→α) (((2ax+b)^2 cos(ax^2 +bx+c))/2)=(((2aα+b)^2 )/2)
$${hop} \\ $$$${lim}_{{x}\rightarrow\alpha} \:\:\frac{\left(\mathrm{2}{ax}+{b}\right){sin}\left({ax}^{\mathrm{2}} +{bx}+{c}\right)}{\mathrm{2}\left({x}−\alpha\right)}\overset{{hop}} {=} \\ $$$${lim}_{{x}\rightarrow\alpha} \:\:\frac{\left(\mathrm{2}{ax}+{b}\right)^{\mathrm{2}} {cos}\left({ax}^{\mathrm{2}} +{bx}+{c}\right)}{\mathrm{2}}=\frac{\left(\mathrm{2}{a}\alpha+{b}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$
Commented by mathmax by abdo last updated on 11/Jul/19
let u(x)=ax^2 +bx +c we have lim_(x→α) u(x)=0 and 1−cos(u(x)) ∼((u^2 (x))/2) (x∈V(α)) ⇒ ((1−cosu(x))/((x−α)^2 )) ∼((u^2 (x))/(2(x−α)^2 )) =(((a(x−α)(x−β))^2 )/(2(x−α)^2 )) =((a^2 (x−β)^2 )/2) ⇒ lim_(x→α) ((1−cosu(x))/((x−α)^2 )) =(a^2 /2)(α−β)^2 .
$${let}\:{u}\left({x}\right)={ax}^{\mathrm{2}} \:+{bx}\:+{c}\:{we}\:{have}\:{lim}_{{x}\rightarrow\alpha} {u}\left({x}\right)=\mathrm{0}\:{and}\: \\ $$$$\mathrm{1}−{cos}\left({u}\left({x}\right)\right)\:\sim\frac{{u}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}}\:\:\:\left({x}\in{V}\left(\alpha\right)\right)\:\:\Rightarrow \\ $$$$\frac{\mathrm{1}−{cosu}\left({x}\right)}{\left({x}−\alpha\right)^{\mathrm{2}} }\:\sim\frac{{u}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}\left({x}−\alpha\right)^{\mathrm{2}} }\:=\frac{\left({a}\left({x}−\alpha\right)\left({x}−\beta\right)\right)^{\mathrm{2}} }{\mathrm{2}\left({x}−\alpha\right)^{\mathrm{2}} }\:=\frac{{a}^{\mathrm{2}} \left({x}−\beta\right)^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\alpha} \:\:\frac{\mathrm{1}−{cosu}\left({x}\right)}{\left({x}−\alpha\right)^{\mathrm{2}} }\:=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\alpha−\beta\right)^{\mathrm{2}} \:. \\ $$
Commented by raj last updated on 11/Jul/19
thank you
$${thank}\:{you} \\ $$
Commented by mathmax by abdo last updated on 12/Jul/19
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$

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