Question

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

Answer 1

$e^{-\frac{a^2}{b^2}}$

Answer 2

The limit is of the indeterminate form $1^\infty$. Using the property $\lim_{x \to c} [f(x)]^{g(x)} = e^{\lim_{x \to c} g(x) [f(x) - 1]}$, we identify $f(x) = \sin^2\left(\frac{\pi}{2-ax}\right)$ and $g(x) = \sec^2\left(\frac{\pi}{2-bx}\right)$. By applying trigonometric identities $\sin(\frac{\pi}{2}-\theta) = \cos\theta$ and $\sec\theta = 1/\cos\theta$, we simplify $f(x)$ to $\cos^2(ax)$ and $g(x)$ to $\frac{1}{\sin^2(bx)}$. Then $f(x)-1 = \cos^2(ax)-1 = -\sin^2(ax)$. The exponent limit becomes $\lim_{x \to 0} \frac{1}{\sin^2(bx)}(-\sin^2(ax)) = \lim_{x \to 0} -\frac{\sin^2(ax)}{\sin^2(bx)}$. Using the standard limit $\lim_{u \to 0} \frac{\sin u}{u}=1$, this simplifies to $-\frac{a^2}{b^2}$. Therefore, the original limit is $e^{-a^2/b^2}$.