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Question Number 183673 by universe last updated on 28/Dec/22
lim_(xβ†’0) [sin^2 ((𝛑/(2βˆ’ax)))]^(sec^2 ((𝛑/(2βˆ’bx))))
$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\:\:\left[\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2}βˆ’\boldsymbol{{a}\mathrm{x}}}\right)\right]^{\boldsymbol{\mathrm{sec}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2}βˆ’\boldsymbol{\mathrm{bx}}}\right)} \\ $$
Answered by AlexandreT last updated on 28/Dec/22
lim [sen^2 ((Ο€/(2βˆ’ax)))βˆ’1]sec^2 ((Ο€/(2βˆ’bx))) xβ†’0 sec^2 ((Ο€/(2βˆ’bx)))=(1/(cos^2 (2βˆ’bx))) βˆ’lim((1βˆ’sen^2 ((Ο€/(2βˆ’ax))))/(1βˆ’sen^2 ((Ο€/(2βˆ’bx))))) xβ†’0 βˆ’lim((1βˆ’((Ο€/(2βˆ’ax)))^2 )/(1βˆ’((Ο€/(2βˆ’bx)))^2 )) xβ†’0 βˆ’lim((((2βˆ’ax)^2 βˆ’Ο€^2 )/((2βˆ’ax)^2 ))/(((2βˆ’bx)^2 βˆ’Ο€^2 )/((2βˆ’bx)^2 ))) xβ†’0 βˆ’lim(((2βˆ’ax)^2 βˆ’Ο€^2 )/((2βˆ’ax)^2 ))Γ—(((2βˆ’bx)^2 )/((2βˆ’bx)^2 βˆ’Ο€^2 )) xβ†’0 =βˆ’[((2^2 βˆ’Ο€^2 )/2^2 )Γ—(2^2 /(2^2 βˆ’Ο€^2 ))] = βˆ’1
$${lim}\:\left[{sen}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}βˆ’{ax}}\right)βˆ’\mathrm{1}\right]{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}βˆ’{bx}}\right) \\ $$$${x}\rightarrow\mathrm{0} \\ $$$${sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}βˆ’{bx}}\right)=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\mathrm{2}βˆ’{bx}\right)} \\ $$$$βˆ’{lim}\frac{\mathrm{1}βˆ’{sen}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}βˆ’{ax}}\right)}{\mathrm{1}βˆ’{sen}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}βˆ’{bx}}\right)} \\ $$$${x}\rightarrow\mathrm{0} \\ $$$$βˆ’{lim}\frac{\mathrm{1}βˆ’\left(\frac{\pi}{\mathrm{2}βˆ’{ax}}\right)^{\mathrm{2}} }{\mathrm{1}βˆ’\left(\frac{\pi}{\mathrm{2}βˆ’{bx}}\right)^{\mathrm{2}} } \\ $$$${x}\rightarrow\mathrm{0} \\ $$$$βˆ’{lim}\frac{\frac{\left(\mathrm{2}βˆ’{ax}\right)^{\mathrm{2}} βˆ’\pi^{\mathrm{2}} }{\left(\mathrm{2}βˆ’{ax}\right)^{\mathrm{2}} }}{\frac{\left(\mathrm{2}βˆ’{bx}\right)^{\mathrm{2}} βˆ’\pi^{\mathrm{2}} }{\left(\mathrm{2}βˆ’{bx}\right)^{\mathrm{2}} }} \\ $$$${x}\rightarrow\mathrm{0} \\ $$$$βˆ’{lim}\frac{\left(\mathrm{2}βˆ’{ax}\right)^{\mathrm{2}} βˆ’\pi^{\mathrm{2}} }{\left(\mathrm{2}βˆ’{ax}\right)^{\mathrm{2}} }Γ—\frac{\left(\mathrm{2}βˆ’{bx}\right)^{\mathrm{2}} }{\left(\mathrm{2}βˆ’{bx}\right)^{\mathrm{2}} βˆ’\pi^{\mathrm{2}} } \\ $$$${x}\rightarrow\mathrm{0} \\ $$$$=βˆ’\left[\frac{\mathrm{2}^{\mathrm{2}} βˆ’\pi^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }Γ—\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} βˆ’\pi^{\mathrm{2}} }\right] \\ $$$$=\:βˆ’\mathrm{1} \\ $$

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