Evaluation of limit at infinity: $\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$

Here, we present a way forward that does not rely on differential calculus. Rather, we use only elementary inequalities and the squeeze theorem. To that end, we proceed.

Inequalities 1:

Recall from basic geometry, that the sine function is bounded as

$$\bbox[5px,border:2px solid #C0A000]{\theta \cos(\theta)\le \sin(\theta)\le \theta} \tag 1$$

for $0\le \theta\le \pi/2$ (SEE THIS ANSWER).

Letting $\theta = \arccos(t)$ in $(1)$ reveals

$$\bbox[5px,border:2px solid #C0A000]{\sqrt{1-t^2}\le \arccos(t)\le \frac{\sqrt{1-t^2}}{t}} \tag 2$$

for $0<t\le 1$.

Inequalities 2:

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality the logarithm function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{t-1}{t}\le \log(t)\le t-1} \tag 3$$

for $t>0$.

Let $L$ be defined as the limit

$$L=\lim_{x\to \infty}x^2\sin\left(\log\left(\cos^{1/2}(\pi/x)\right)\right)$$

Enforce the substitution $x=\frac{1}{\pi}\arccos(t)$. Then, we can express the limit $L$ as

$$L=\lim_{t\to 1^-}\pi^2 \frac{\sin\left(\frac12\log\left(t\right)\right)}{\arccos^2(t)}$$

Applying inequalities $(1)-(3)$ we find that

$$\pi^2 \left(\frac{\frac12 \cos(\log(t))\,\frac{t-1}{t}}{\frac{1-t^2}{t^2}}\right)\le \pi^2 \frac{\sin\left(\frac12\log\left(t\right)\right)}{\arccos^2(t)}\le \pi^2 \left(\frac{\frac12 (t-1)}{(1-t^2)}\right)$$

or after simplifying

$$-\left(\frac{\pi^2}{2(1+t)}\right)t\,\cos(\log(t))\le \pi^2 \frac{\sin\left(\frac12\log\left(t\right)\right)}{\arccos^2(t)}\le -\left(\frac{\pi^2}{2(1+t)} \right) \tag 4$$

Applying the squeeze theorem to $(4)$ yields the coveted equality

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}x^2\sin\left(\log\left(\cos^{1/2}(\pi/x)\right)\right)=-\frac{\pi^2}{4}}$$

Using Taylor series for $\cos u$, $\sqrt{1+u}$, $\ln(1+u)$, and $\sin u$ when $u\to 0$. When $x\to \infty$, $\frac{1}{x} \to 0$, so

$$\cos\frac{\pi}{x} = 1- \frac{\pi^2}{2x^2} + o\left(\frac{1}{x^2}\right)$$ and $$\sqrt{ \cos(\frac{\pi}{x}) } = \sqrt{ 1- \frac{\pi^2}{2x^2} + o\left(\frac{1}{x^2}\right) } = 1- \frac{\pi^2}{4x^2} + o\left(\frac{1}{x^2}\right)$$ so that $$\ln \sqrt{ \cos(\frac{\pi}{x}) } = \ln\left (1- \frac{\pi^2}{4x^2} + o\left(\frac{1}{x^2}\right) \right) = - \frac{\pi^2}{4x^2} + o\left(\frac{1}{x^2}\right)$$ and finally $$ x^2 \sin \ln \sqrt{ \cos(\frac{\pi}{x}) } = x^2 \sin\!\left( - \frac{\pi^2}{4x^2} + o\!\left(\frac{1}{x^2}\right)\right) = -x^2 \left(\frac{\pi^2}{4x^2} + o\!\left(\frac{1}{x^2}\right)\right) = - \frac{\pi^2}{4} + o(1)$$