Evaluate $\lim_{x\to 0} [\sin^2(\frac{\pi}{2-px})]^{\sec^2(\frac{\pi}{2-qx})}$.

Question

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.

You've dropped a sign, as $\sin^2y-1=-\cos^2y$. – J.G. Feb 09 '23 at 13:12 — J.G., Feb 09 '23 at 13:12
@J.G. Thank you. I have fixed it. – StrawberryVanilla Feb 09 '23 at 13:16 — StrawberryVanilla, Feb 09 '23 at 13:16

score 2 · Answer 1 · answered Feb 09 '23 at 13:22

2

As $x\to0,$ $$\cos\frac{\pi}{2-px}=\sin\left(\frac\pi2-\frac\pi{2-px}\right)=\sin\frac{-\pi px}{2(2-px)}\sim-\frac{\pi px}4$$ hence $$\frac{-\cos^2\frac{\pi}{2-px}}{\cos^2\frac{\pi}{2-qx}}\to-\frac{p^2}{q^2}.$$

answered Feb 09 '23 at 13:22

Anne Bauval

34,650

score -1 · Answer 2 · answered Feb 09 '23 at 13:42

$$\lim_{x\to0}\sin^2\left(\dfrac\pi{2-px}\right)^{\sec^2\left(\dfrac\pi{2-px}\right)}$$

$$=\left(\lim_{x\to0}\left(1-\cos^2\left(\dfrac\pi{2-px}\right)\right)^{-\dfrac1{\cos^2\left(\dfrac\pi{2-px}\right)}}\right)^{-\left(\lim_{x\to0}\dfrac{\cos\dfrac\pi{2-px}}{\cos\dfrac\pi{2-qx}}\right)^2}$$

The inner limit converges to $?$

Finally, for the exponent using $\cos\left(\dfrac\pi2-y\right)=\sin y,$ $\lim_{x\to0}\dfrac{\cos\dfrac\pi{2-px}}{\cos\dfrac\pi{2-qx}}=\lim_{x\to0}\dfrac{-\sin\dfrac{\pi p x}{2-px}}{-\sin\dfrac{\pi q x}{2-qx}}=\lim_{x\to0}\dfrac{\dfrac{\pi p x}{2-px}}{\dfrac{\pi q x}{2-qx}}=\dfrac pq$

score -1 · Accepted Answer · answered Feb 09 '23 at 14:00

$$ \begin{align} L & = \lim_{x \to 0} - \frac{\cos^2(\frac{\pi}{2-px})}{\cos^2(\frac{\pi}{2-qx})} \\ & = \lim_{x \to 0} - \frac{\sin^2(\frac\pi2-\frac{\pi}{2-px})}{\sin^2(\frac\pi2-\frac{\pi}{2-qx})} \\ & = \lim_{x \to 0} - \left(\frac{\sin\left(\frac\pi2-\frac{\pi}{2-px}\right)}{\frac\pi2-\frac{\pi}{2-px}}\right)^2\left(\frac{\frac\pi2-\frac{\pi}{2-qx}}{\sin^2\left(\frac\pi2-\frac{\pi}{2-qx}\right)}\right)^2\left(\frac{\frac\pi2-\frac{\pi}{2-px}}{\frac\pi2-\frac{\pi}{2-qx}}\right)^2 \\ & = \lim_{x \to 0} - \left(\frac{\sin\left(-\frac{\pi p x}{2(2-px)}\right)}{-\frac{\pi p x}{2(2-px)}}\right)^2\left(\frac{-\frac{\pi qx}{2(2-qx)}}{\sin^2\left(-\frac{\pi qx}{2(2-qx)}\right)}\right)^2\left(\frac{p(2-qx)}{q(2-px)}\right)^2 \\ & = -\frac{p^2}{q^2} \end{align} $$

So, $$\lim_{x\to 0} \left(\sin^2(\frac{\pi}{2-px})\right)^{\sec^2(\frac{\pi}{2-qx})} = \large e^{\lim_{x\to 0} \frac{-\cos^2(\frac{\pi}{2-px})}{\cos^2(\frac{\pi}{2-qx})}} = \large e^{-\frac{p^2}{q^2}}$$

Evaluate $\lim_{x\to 0} [\sin^2(\frac{\pi}{2-px})]^{\sec^2(\frac{\pi}{2-qx})}$.

3 Answers3