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

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 $-\pi^2/4$? 
