Question

Solve: $\underset{x\to 0}{lim}{\sin }^2{\left(\frac{\pi }{2-ax}\right)}^{{\sec }^2\frac{\pi }{2-bx}}$

Answer 1

We have, $\underset{x\to 0}{lim}{\sin }^2{\left(\frac{\pi }{2-ax}\right)}^{{\sec }^2\frac{\pi }{2-bx}}$

$=$$e^{-\underset{x\to 0}{lim}\frac{co{s}^2\left(\pi /\right(2-ax\left)\right)}{co{s}^2(\pi /2-bx)}}$

$=e^{-\underset{x\to 0}{lim}\frac{si{n}^2(\pi /2-\pi /(2-ax\left)\right)}{si{n}^2(\pi /2-\pi /(2-bx\left)\right)}}$

$=e^{-\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi ax/2\right(2-ax\left)\right)}{si{n}^2\left(\pi bx/2\right(2-bx\left)\right)}}=e^{-\frac{4a^2}{4b^2}}=e^{-a^2/b^2}$