How can I calculate $\displaystyle\lim_{x \to \infty}x^{2}\ln\left(\cos \left(\pi/x\right)\right) $? Does anybody know how to solve this?
$$\lim_{x \to \infty}x^{2}\ln\left(\cos\left(\pi \over x\right)\right)$$ 
 A: Hint:  as $x \to \infty, \frac \pi x \to 0$  Use the Taylor series of $\cos$ and use that with one you know of $\log$
A: We have
$$\ln(\cos(\pi/x)) = \dfrac12\ln(\cos^2(\pi/x)) = \dfrac12\ln(1-\sin^2(\pi/x)) = -\dfrac12 \left( \sum_{k=1}^{\infty} \dfrac{\sin^{2k}(\pi/x)}k\right)$$
Hence,
$$x^2\ln(\cos(\pi/x)) = -\dfrac12 \left( \sum_{k=1}^{\infty} x^2\dfrac{\sin^{2k}(\pi/x)}k\right)$$
Now take limit as $x\to \infty$ and use the fact that
$$\lim_{x \to \infty} x^2 \sin^{2k}(\pi/x) = \begin{cases} \pi^2 & k=1\\ 0 & k>1 \end{cases}$$
A: $\newcommand{\+}{^{\dagger}}%
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$$
x^{2}\ln\pars{\cos\pars{\pi \over x}}
\sim
x^{2}\ln\pars{1 - {\pi^{2} \over 2x^{2}}}
\sim
x^{2}\pars{-\,{\pi^{2} \over 2x^{2}}}\quad\mbox{when}\quad\verts{x} \gg 1
$$
A: As $x\to\infty,\,\pi/x\to0$; so $\cos\left(\frac{\pi}{x}\right)\sim 1-\frac{\pi^2}{2x^2}$ and $\ln\left(1-\frac{\pi^2}{2x^2}\right)\sim \left(-\frac{\pi^2}{2x^2}\right)$ so for $x\to\infty$
$$
x^2\ln\cos\left(\frac{\pi}{x}\right)\sim x^2\ln\left(1-\frac{\pi^2}{2x^2}\right)\sim x^2\left(-\frac{\pi^2}{2x^2}\right)=-\frac{\pi^2}{2}
$$
A: Let $x=1/t$. So we are interested in the limit as $t\to 0^+$ of $\ln(\cos(\pi t))/t^2$.
One round of L'Hospital's Rule brings us to 
$$\lim_{t\to 0^+} -\frac{\pi}{2\cos(\pi t)} \frac{\sin(\pi t)}{t}.$$
The first part is nicely behaved near $0$. For $\lim_{t\to 0^+} \frac{\sin(\pi t)}{t}$, use L'Hospital's Rule, or simpler tools.
A: Hint: Use a change of variables $u = \frac{\pi}{x}$ and then it will be solvable by first year calculus (you do not even have to use a Taylor's series).
A: An entirely different approach is to reexpess $x^2 \ln{\big(\cos{(\frac{\pi}{x})}\big)}$ as  $\ln{\big(\big(\cos{(\frac{\pi}{x}})\big)^{x^2}\big)}$.
