<s>[INST] Evaluate $\lim_{x\to 0} [\sin^2(\frac{\pi}{2-px})]^{\sec^2(\frac{\pi}{2-qx})}$. I am attempting to evaluate
$\lim_{x\to 0} [\sin^2(\frac{\pi}{2-px})]^{\sec^2(\frac{\pi}{2-qx})}$
where $p,q\in \Re$
This is the $1^\infty$ form.
We have a general formula for this indeterminate form-
$\lim_{x\to a}f^g$ where $f\to 1$ and $g\to \infty$ is equal to $e^{\lim_{x\to a}g(f-1)}$
On using the formula, the problem becomes
$\large e^{\lim_{x\to 0} [\sin^2(\frac{\pi}{2-px})-1]{\sec^2(\frac{\pi}{2-qx})}}$
$\large e^{\lim_{x\to 0} [-\cos^2(\frac{\pi}{2-px})]{\sec^2(\frac{\pi}{2-qx})}}$
$\large e^{\lim_{x\to 0} \frac{-\cos^2(\frac{\pi}{2-px})}{\cos^2(\frac{\pi}{2-qx})}}$
How do I proceed further? Any other methods to solve such a problem are welcome.
 [/INST] $$ \begin{align}
 L & = \lim_{x \to 0} - \frac{\cos^2(\frac{\pi}{2-px})}{\cos^2(\frac{\pi}{2-qx})} \\ 
& = \lim_{x \to 0} - \frac{\sin^2(\frac\pi2-\frac{\pi}{2-px})}{\sin^2(\frac\pi2-\frac{\pi}{2-qx})} \\
& = \lim_{x \to 0} - \left(\frac{\sin\left(\frac\pi2-\frac{\pi}{2-px}\right)}{\frac\pi2-\frac{\pi}{2-px}}\right)^2\left(\frac{\frac\pi2-\frac{\pi}{2-qx}}{\sin^2\left(\frac\pi2-\frac{\pi}{2-qx}\right)}\right)^2\left(\frac{\frac\pi2-\frac{\pi}{2-px}}{\frac\pi2-\frac{\pi}{2-qx}}\right)^2 \\
& = \lim_{x \to 0} - \left(\frac{\sin\left(-\frac{\pi p x}{2(2-px)}\right)}{-\frac{\pi p x}{2(2-px)}}\right)^2\left(\frac{-\frac{\pi qx}{2(2-qx)}}{\sin^2\left(-\frac{\pi qx}{2(2-qx)}\right)}\right)^2\left(\frac{p(2-qx)}{q(2-px)}\right)^2 \\
& = -\frac{p^2}{q^2}
\end{align} $$
So,
$$\lim_{x\to 0} \left(\sin^2(\frac{\pi}{2-px})\right)^{\sec^2(\frac{\pi}{2-qx})} = \large e^{\lim_{x\to 0} \frac{-\cos^2(\frac{\pi}{2-px})}{\cos^2(\frac{\pi}{2-qx})}} = \large e^{-\frac{p^2}{q^2}}$$
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