lim-x-0-sin-2-2-ax-sec-2-2-bx-

Question Number 183673 by universe last updated on 28/Dec/22

$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\:\:\left[\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2}−\boldsymbol{{a}\mathrm{x}}}\right)\right]^{\boldsymbol{\mathrm{sec}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2}−\boldsymbol{\mathrm{bx}}}\right)} \\ $$

Answered by AlexandreT last updated on 28/Dec/22

$${lim}\:\left[{sen}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}−{ax}}\right)−\mathrm{1}\right]{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}−{bx}}\right) \\ $$$${x}\rightarrow\mathrm{0} \\ $$$${sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}−{bx}}\right)=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\mathrm{2}−{bx}\right)} \\ $$$$−{lim}\frac{\mathrm{1}−{sen}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}−{ax}}\right)}{\mathrm{1}−{sen}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}−{bx}}\right)} \\ $$$${x}\rightarrow\mathrm{0} \\ $$$$−{lim}\frac{\mathrm{1}−\left(\frac{\pi}{\mathrm{2}−{ax}}\right)^{\mathrm{2}} }{\mathrm{1}−\left(\frac{\pi}{\mathrm{2}−{bx}}\right)^{\mathrm{2}} } \\ $$$${x}\rightarrow\mathrm{0} \\ $$$$−{lim}\frac{\frac{\left(\mathrm{2}−{ax}\right)^{\mathrm{2}} −\pi^{\mathrm{2}} }{\left(\mathrm{2}−{ax}\right)^{\mathrm{2}} }}{\frac{\left(\mathrm{2}−{bx}\right)^{\mathrm{2}} −\pi^{\mathrm{2}} }{\left(\mathrm{2}−{bx}\right)^{\mathrm{2}} }} \\ $$$${x}\rightarrow\mathrm{0} \\ $$$$−{lim}\frac{\left(\mathrm{2}−{ax}\right)^{\mathrm{2}} −\pi^{\mathrm{2}} }{\left(\mathrm{2}−{ax}\right)^{\mathrm{2}} }×\frac{\left(\mathrm{2}−{bx}\right)^{\mathrm{2}} }{\left(\mathrm{2}−{bx}\right)^{\mathrm{2}} −\pi^{\mathrm{2}} } \\ $$$${x}\rightarrow\mathrm{0} \\ $$$$=−\left[\frac{\mathrm{2}^{\mathrm{2}} −\pi^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} −\pi^{\mathrm{2}} }\right] \\ $$$$=\:−\mathrm{1} \\ $$

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