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?
$$\lim_{x\rightarrow\infty}\left(\cos\sqrt{\frac{2\pi}{x}}\right)^x$$

$$\cos\sqrt{\frac{2\pi}x}=1-2\sin^2\sqrt{\frac{\pi}{2x}}$$
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
$$e^{-\pi}.$$
