Evaluation of limit at infinity: $\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$ $$\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$$
What I tried was writing $1/x=t$ and making the limit tend to zero and writing the cos term in the form of sin
 A: 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
$$
x^2 \sin \ln \sqrt{ \cos(\frac{\pi}{x}) } = x^2 \sin\!\left( - \frac{\pi^2}{4x^2} + o\!\left(\frac{1}{x^2}\right)\right)
= -x^2 \left(\frac{\pi^2}{4x^2} + o\!\left(\frac{1}{x^2}\right)\right) = - \frac{\pi^2}{4} + o(1)$$
A: $\ln(\cos(\pi /x)^{1/2}) = 1/4\, \ln(\cos (\pi /x)^2) = 1/4\, \ln(1 - (\sin (\pi /x))^2)$. Also, $\lim_{z \to 0}\frac {\sin z} z = 1$ and $\lim_{u \to 0 }\frac {\ln(1-u)} u = -1$ (for example by l'Hopital). Now:
$$x^2 \sin(\ln(\cos(\pi/x)^{1/2})) = \\[12pt]  x^2\cdot \frac {\sin(\ln(\cos(\pi/x)^{1/2}))} {\ln(\cos(\pi/x)^{1/2}))} \cdot \frac {1/4 \,\ln(1 - (\sin(\pi/x)^2))} {(\sin(\pi /x))^2}\cdot \frac{(\sin(\pi/ x))^2}{(\pi/ x)^2}\cdot (\pi/x)^2
$$
Combine the first and last factor to get $\pi^2$, the second and fourth factor both have limit 1, and the middle factor has limit $-1/4$. So the overall answer is $-\pi^2/4$.
Many similar questions can be reduced to fraction algebra and common limits, without using Taylor series.
A: Notice that
\begin{align*}
\ln\big(\cos(\pi/x)^{1/2}\big) &= \frac 1 2 \ln \cos \frac{\pi}{x} \\
&= \frac 1 2 \ln \left(1 - \frac 1 2\frac{\pi^2}{x^2} + O\left(\frac 1 {x^4}\right)\right) \\
&= -\frac 1 4 \frac{\pi^2}{x^2} + O \left( \frac 1 {x^4}\right)
\end{align*}
Now do you see how to show that the desired limit is $-\pi^2/4$? 
A: Push $x^2$ to the bottom, then apply L'Hospital rule.
$\lim_{x\to \infty} \frac{\sin (\frac{\ln (\cos (\pi/x)}{2})}{x^{-2}}$ \ applying L'Hospital once and some simplification you will get, \
$-\pi^2/4 \lim_{x\to \infty} [\frac{\tan (\pi/x)}{\pi/x} \cos (\frac{\ln (\cos (\pi/x)}{2})]=-\pi^2/4$.
A: Using $\pi/x=t$, you can compute
$$
\lim_{t\to0^+}\frac{\pi^2}{t^2}\sin\left(\frac{1}{2}\ln\cos t\right)
$$
Pulling the exponent out of the logarithm is possible because $\cos t$ is positive in a neighborhood of $0$.
Now recall that
$$
\cos t=1-\frac{t^2}{2}+o(t^2)
$$
so
$$
\ln(\cos t)=-\frac{t^2}{2}+o(t^2)
$$
and so your limit is
$$
\lim_{t\to0^+}-\frac{\pi^2}{4}\frac{t^2+o(t^2)}{t^2}
$$
A: 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
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}x^2\sin\left(\log\left(\cos^{1/2}(\pi/x)\right)\right)=-\frac{\pi^2}{4}}$$
A: By Taylor limited to the useful terms ($\cos(t)\approx1-\dfrac{t^2}2$, $\ln(1+t)\approx t$, $\sin(t)\approx t$),
$$\cos\left(\frac\pi x\right)\approx1-\frac{\pi^2}{2x^2},$$
$$\ln\left(\left(1-\frac{\pi^2}{2x^2}\right)^{1/2}\right)\approx-\frac12\frac{\pi^2}{2x^2},$$
$$-\sin\left(\frac{\pi^2}{4x^2}\right)\approx-\frac{\pi^2}{4x^2}.$$
Hence $$-\frac{\pi^2}4.$$

But how do we known how many terms to keep ?
The cosine gives even terms, with the constant coefficient $1$. The logarithm suppresses the coefficient $1$ and leaves the next term unchanged. The sine also leaves this term unchanged.
