Computing $\lim\limits_{x \to \infty} \left( \cos\sqrt{ {2 \pi \over x}} \right)^x$ I have tried to solve the below limit using the exponential formula and then applying l'Hospital but the problem turns hard to solve. Does anyone knows an easier way for it?
$$\lim_{x\rightarrow\infty}\left(\cos\sqrt{\frac{2\pi}{x}}\right)^x$$
 A: Let $$f(x) = x\log\left(\cos\sqrt{\frac{2\pi}{x}}\right)
$$
so that your expression is 
$$
\lim_{x\to\infty} e^{f(x)} = e^{\lim_{x\to\infty} f(x)}
$$
Now let $u=1/x$ so that we need to find
$$
\lim_{u\to 0} \frac{\log\left(\cos\sqrt{2\pi u}\right)}{u}
$$
Now apply l'Hopital and the fact that $\lim_{z\to 0} \tan(az)/z = a$
to get
$$
\lim_{u\to 0} \frac{\log\left(\cos\sqrt{2\pi u}\right)}{u}
=-\lim_{u\to 0} \frac{\sqrt{\frac\pi2}\tan\sqrt{2\pi u}/\sqrt{u}}{1}=-\pi\sqrt{\frac\pi2}\sqrt{2\pi}
$$
giving a result for the original problem of 
$$
e^{-\pi}
$$
A: Stop thinking L'Hospital's rule is the panacea for determining limits! Here you simply need Taylor's expansion at order $2$.
First, it is enough to have the limit of the log, $\; x\ln\biggl(\cos\sqrt{\dfrac{2\pi} x}\biggr)$. So set $u= \sqrt{\dfrac{2\pi} x}$: $u$ tends to $0$ when $x$ tends to $\infty$, and
\begin{align} x\ln\biggl(\cos\sqrt{\dfrac{2\pi} x}\biggr)&= \dfrac{2\pi}{u^2}\ln(\cos u)=\dfrac{2\pi}{u^2}\ln\Bigl(1-\dfrac{u^2}2+o(u^2)\Bigr)\\&=\dfrac{2\pi}{u^2}\Bigl(-\dfrac{u^2}2+o(u^2)\Bigr)=-\pi +o(1)\to -\pi
\end{align}
Hence the limit we seek is equal to $\mathrm e^{-\pi}$.
A: $$\cos\sqrt{\frac{2\pi}x}=1-2\sin^2\sqrt{\frac{\pi}{2x}}$$
and
$$x=\frac{\pi\left(\dfrac{\sin\sqrt{\dfrac{\pi}{2x}}}{\sqrt{\dfrac\pi{2x}}}\right)^2}{2\sin^2\sqrt{\dfrac{\pi}{2x}}}.$$
The quotient inside the parenthesis tends to one, and the limit is that of
$$\left(\left(1-2\sin^2\sqrt{\frac{\pi}{2x}}\right)^{1/2\sin^2\sqrt{\frac{\pi}{2x}}}\right)^{\pi(\cdots)^2}$$
or
$$e^{-\pi}.$$
