TITLE: Limit of $x^2\sin\left(\ln\sqrt{\cos\frac{\pi}{x}}\right)$
QUESTION [10 upvotes]: Find $$\lim_{x\to\infty}x^2\sin\left(\ln\sqrt{\cos\frac{\pi}{x}}\right).$$
I tried substituting $x=1/t$ with $t$ approaching $0$ but the term inside the bracket is not giving me ideas on how to compute the limit.

REPLY [3 votes]: Using $\lim\limits_{t\to0}\frac{\sin(t)}t=1$ and $\lim\limits_{t\to0}\frac{\log(1-t)}t=-1$, we get
$$
\begin{align}
\lim_{x\to\infty}x^2\sin\left(\log\left(\sqrt{\cos(\pi/x)}\right)\right)
&=\pi^2\lim_{t\to0}\frac{\sin\left(\frac12\log(\cos(t))\right)}{t^2}\\
&=\pi^2\lim_{t\to0}\frac{\frac12\log(\cos(t))}{t^2}\frac{\sin\left(\frac12\log(\cos(t))\right)}{\frac12\log(\cos(t))}\\
&=\pi^2\lim_{t\to0}\frac{\frac14\log(\cos^2(t))}{t^2}\cdot1\\
&=\frac{\pi^2}4\lim_{t\to0}\frac{\log\left(1-\sin^2(t)\right)}{t^2}\\
&=\frac{\pi^2}4\lim_{t\to0}\frac{\sin^2(t)}{t^2}\frac{\log\left(1-\sin^2(t)\right)}{\sin^2(t)}\\
&=\frac{\pi^2}4\left(\lim_{t\to0}\frac{\sin(t)}{t}\right)^2\cdot(-1)\\[4pt]
&=-\frac{\pi^2}4
\end{align}
$$