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

$$\underset{x\to 0}{\mathbf{lim}}{\left[{\mathbf{sin}}^2\left(\frac{\mathit{\pi }}{2-\mathit{a}\mathbf{x}}\right)\right]}^{{\mathbf{sec}}^2\left(\frac{\mathit{\pi }}{2-\mathbf{bx}}\right)}$$

Answered by AlexandreT last updated on 28/Dec/22

$$lim[{sen}^2\left(\frac{\pi }{2-ax}\right)-1]{sec}^2\left(\frac{\pi }{2-bx}\right)$$$$x\to 0$$$${sec}^2\left(\frac{\pi }{2-bx}\right)=\frac{1}{{cos}^2(2-bx)}$$$$-lim\frac{1-{sen}^2\left(\frac{\pi }{2-ax}\right)}{1-{sen}^2\left(\frac{\pi }{2-bx}\right)}$$$$x\to 0$$$$-lim\frac{1-{\left(\frac{\pi }{2-ax}\right)}^2}{1-{\left(\frac{\pi }{2-bx}\right)}^2}$$$$x\to 0$$$$-lim\frac{\frac{{(2-ax)}^2-{\pi }^2}{{(2-ax)}^2}}{\frac{{(2-bx)}^2-{\pi }^2}{{(2-bx)}^2}}$$$$x\to 0$$$$-lim\frac{{(2-ax)}^2-{\pi }^2}{{(2-ax)}^2}\times \frac{{(2-bx)}^2}{{(2-bx)}^2-{\pi }^2}$$$$x\to 0$$$$=-\left[\frac{2^2-{\pi }^2}{2^2}\times \frac{2^2}{2^2-{\pi }^2}\right]$$$$=-1$$

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