Limit of $x^2\sin\left(\ln\sqrt{\cos\frac{\pi}{x}}\right)$ 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.
 A: Recall that, as $u \to 0$, we have
$$
\begin{align}
\cos u& =1-\frac {u^2}{2}+\mathcal{O}(u^3)\\
\sin u& =u+\mathcal{O}(u^3)\\
\ln (1+u)&=u-\frac {u^2}{2}+\mathcal{O}(u^3)
\end{align}
$$ giving, as $x \to \infty$,
$$
\cos\frac{\pi}{x}=1-\frac{\pi^2}{2x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)
$$ 
$$
\begin{align}
\log \left(\cos\frac{\pi}{x}\right)&=\log \left(1-\frac{\pi^2}{2x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)\right)\\\\
\log \left(\cos\frac{\pi}{x}\right)&=-\frac{\pi^2}{2x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)\\\\
\log \left(\sqrt{\cos\frac{\pi}{x}}\right)&=-\frac{\pi^2}{4x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)
\end{align}
$$ and
$$
\begin{align}
\sin \left(\log \left(\sqrt{\cos\frac{\pi}{x}}\right)\right)&=-\frac{\pi^2}{4x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)\\\\
x^2\sin \left(\log \left(\sqrt{\cos\frac{\pi}{x}}\right)\right)&=-\frac{\pi^2}{4}+\mathcal{O}\left(\frac{1}{x}\right)
\end{align}
$$ giving $-\dfrac{\pi^2}{4}$ for the desired limit.
A: Do the substitution $t=1/x$, so the limit is easier:
$$
\lim_{t\to0^+}\frac{\sin\ln\sqrt{\cos(\pi t)}}{t^2}
$$
Now observe that, since $\ln\sqrt{\cos\pi t}$ is monotonic in a right neighborhood of $0$, you can say that
$$
\lim_{t\to0^+}\frac{\sin\ln\sqrt{\cos(\pi t)}}{\ln\sqrt{\cos(\pi t)}}=1
$$
so you want to compute
$$
\lim_{t\to0^+}\frac{\ln\sqrt{\cos(\pi t)}}{t^2}=
\frac{1}{2}\lim_{t\to0^+}\frac{\ln\cos(\pi t)}{t^2}
$$
Let's set $\pi t=2u$, so the limit becomes
$$
\frac{\pi^2}{8}\lim_{u\to0^+}\frac{\ln(1-2\sin^2u)}{u^2}
$$
Now recall that $\lim_{z\to0}(\ln(1-z))/z=-1$ and you can write the limit as
$$
-\frac{\pi^2}{8}\lim_{u\to0^+}\frac{2\sin^2u}{u^2}
$$
so the given limit ends up to be $-\pi^2/4$.
A: Using $\lim\limits_{t\to0}\frac{\sin(t)}t=1$ and $\lim\limits_{t\to0}\frac{\log(1-t)}t=-1$, we get
$$
\begin{align}
\lim_{x\to\infty}x^2\sin\left(\log\left(\sqrt{\cos(\pi/x)}\right)\right)
&=\pi^2\lim_{t\to0}\frac{\sin\left(\frac12\log(\cos(t))\right)}{t^2}\\
&=\pi^2\lim_{t\to0}\frac{\frac12\log(\cos(t))}{t^2}\frac{\sin\left(\frac12\log(\cos(t))\right)}{\frac12\log(\cos(t))}\\
&=\pi^2\lim_{t\to0}\frac{\frac14\log(\cos^2(t))}{t^2}\cdot1\\
&=\frac{\pi^2}4\lim_{t\to0}\frac{\log\left(1-\sin^2(t)\right)}{t^2}\\
&=\frac{\pi^2}4\lim_{t\to0}\frac{\sin^2(t)}{t^2}\frac{\log\left(1-\sin^2(t)\right)}{\sin^2(t)}\\
&=\frac{\pi^2}4\left(\lim_{t\to0}\frac{\sin(t)}{t}\right)^2\cdot(-1)\\[4pt]
&=-\frac{\pi^2}4
\end{align}
$$
