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Question Number 82448 by zainal tanjung last updated on 21/Feb/20

Lim_(x→(π/2))  {((sin (6x−3π)^2 −sin (6x−3π)sin (4x−2π))/(5x^2  cos (5x−((5π)/2) )))

$$\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{Lim}}\:\left\{\frac{\mathrm{sin}\:\left(\mathrm{6x}−\mathrm{3}\pi\right)^{\mathrm{2}} −\mathrm{sin}\:\left(\mathrm{6x}−\mathrm{3}\pi\right)\mathrm{sin}\:\left(\mathrm{4x}−\mathrm{2}\pi\right)}{\mathrm{5x}^{\mathrm{2}} \:\mathrm{cos}\:\left(\mathrm{5x}−\frac{\mathrm{5}\pi}{\mathrm{2}}\:\right)}\right. \\ $$

Commented by jagoll last updated on 21/Feb/20

let x−(π/2) = u  lim_(u→0)  (1/(5(u+(π/2))^2 )) × lim_(u→0)  ((sin (6u)^2 −sin 6u sin 4u)/(cos 5u))  = (4/(5π^2 )) × 0 = 0

$${let}\:{x}−\frac{\pi}{\mathrm{2}}\:=\:{u} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{5}\left({u}+\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} }\:×\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{6}{u}\right)^{\mathrm{2}} −\mathrm{sin}\:\mathrm{6}{u}\:\mathrm{sin}\:\mathrm{4}{u}}{\mathrm{cos}\:\mathrm{5}{u}} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{5}\pi^{\mathrm{2}} }\:×\:\mathrm{0}\:=\:\mathrm{0} \\ $$

Commented by zainal tanjung last updated on 21/Feb/20

You are right sir....! Okey....

$$\mathrm{You}\:\mathrm{are}\:\mathrm{right}\:\mathrm{sir}....!\:\mathrm{Okey}.... \\ $$

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