Question:

Step 1: Expand $\sin x$ As $x \to 0$, the Taylor expansion for $\sin x$ is given by $$ \sin x = x - \frac{x^3}{6} + O(x^5) $$ Step 2: Expand $\tan(\sin x)$ Let $u = \sin x$. As $x \to 0$, $u \to 0$. The Taylor expansion for $\tan u$ is given by $$ \tan u = u + \frac{u^3}{3} + O(u^5) $$ Substitute the expansion for $\sin x$: $$ \tan(\sin x) = \left(x - \frac{x^3}{6} + O(x^5)\right) + \frac{\left(x - \frac{x^3}{6} + O(x^5)\right)^3}{3} + O(x^5) $$ $$ \tan(\sin x) = x - \frac{x^3}{6} + \frac{x^3}{3} + O(x^5) = x + \frac{x^3}{6} + O(x^5) $$ Step 3: Expand $\cos^2(\tan(\sin x))$ Let $\theta = \tan(\sin x)$. From Step 2, $\theta = x + \frac{x^3}{6} + O(x^5)$. Use the trigonometric identity $\cos^2\theta = \frac{1+\cos(2\theta)}{2}$. First, expand $2\theta$: $$ 2\theta = 2\left(x + \frac{x^3}{6} + O(x^5)\right) = 2x + \frac{x^3}{3} + O(x^5) $$ Next, expand $\cos(2\theta)$. For small $Z$, $\cos Z = 1 - \frac{Z^2}{2!} + O(Z^4)$. Using the approximation $\cos(2\theta) \approx 1 - x^2 + O(x^4)$ for $2\theta \approx 2x$: $$ \cos(2\theta) = 1 - x^2 + O(x^4) $$ Substitute this into the identity for $\cos^2\theta$: $$ \cos^2(\tan(\sin x)) = \frac{1 + (1 - x^2 + O(x^4))}{2} = \frac{2 - x^2 + O(x^4)}{2} = 1 - \frac{x^2}{2} + O(x^4) $$ Step 4: Expand $\pi \cos^2(\tan(\sin x))$ Multiply the expansion from Step 3 by $\pi$: $$ \pi \cos^2(\tan(\sin x)) = \pi \left(1 - \frac{x^2}{2} + O(x^4)\right) = \pi - \frac{\pi x^2}{2} + O(x^4) $$ Step 5: Expand the outer sine Let $A = \pi \cos^2(\tan(\sin x))$. From Step 4, $A = \pi - \frac{\pi x^2}{2} + O(x^4)$. We need to find $\sin(A)$. Use the identity $\sin(\pi - \phi) = \sin \phi$: $$ \sin(A) = \sin\left(\pi - \left(\frac{\pi x^2}{2} - O(x^4)\right)\right) = \sin\left(\frac{\pi x^2}{2} - O(x^4)\right) $$ For small arguments $Z$, $\sin Z = Z - \frac{Z^3}{6} + O(Z^5)$. $$ \sin(A) = \left(\frac{\pi x^2}{2} - O(x^4)\right) - \frac{\left(\frac{\pi x^2}{2} - O(x^4)\right)^3}{6} + O(x^8) $$ $$ \sin(A) = \frac{\pi x^2}{2} + O(x^4) $$ Step 6: Calculate the limit Substitute the expansion from Step 5 into the limit expression: $$ \lim_{x \to 0} \frac{\sin\left(\pi \cos^2(\tan(\sin x))\right)}{x^2} = \lim_{x \to 0} \frac{\frac{\pi x^2}{2} + O(x^4)}{x^2} $$ $$ = \lim_{x \to 0} \left(\frac{\pi}{2} + O(x^2)\right) = \frac{\pi}{2} $$