An Unfamiliar Limit

$\begin{aligned} A & = \lim_{n \to \infty} \cos^n \frac \pi {\sqrt n} & \small \color{#3D99F6} \text{A }1^\infty \text{ case, the following applies} \\ & = \exp \left(\lim_{n \to \infty} n \left({\color{#3D99F6} \cos \frac \pi {\sqrt n}} - 1\right)\right) & \small \color{#3D99F6} \lim_{x \to \infty} f(x)^{h(x)} = e^{\lim_{n \to \infty}h(x)(f(x)-1)} \\ & = \exp \left(\lim_{n \to \infty} n \left({\color{#3D99F6} 1 - \frac {\pi^2}{2!n} + \frac {\pi^4}{4!n^2} - \cdots } - 1\right)\right) & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \exp \left(\lim_{n \to \infty} \left(- \frac {\pi^2}2 + \frac {\pi^4}{24n} - \frac {\pi^6}{720n^2} + \cdots\right)\right) \\ & = e^{-\frac {\pi^2}2} \approx 0.007192 \end{aligned}$

Therefore, $\lfloor 1000A\rfloor = \boxed 7$ .

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Reference: The $\color{#3D99F6}1^\infty$ case (2nd method)

This is not very formal, but here we go.

When $x \to 0$ , $\cos{x} \to \left( 1-\frac{x^2}{2}\right)$ , so

$\displaystyle \lim_{n \to \infty} \left(\cos{\frac{\pi}{\sqrt{n}}}\right)^n = \lim_{n \to \infty} \left( 1-\frac{{\pi}^2}{2n} \right)^n$

If $\displaystyle p= -\frac{2n}{\pi^2}$ , we have

$\large \displaystyle \lim_{p \to -\infty} \left( 1+\frac{1}{p} \right)^{-\frac{p \pi^2}{2}} = \sqrt{e^{-\pi^2}} = 0.007191...$

So $\lfloor 1000A \rfloor = 7$

$\begin{aligned} A &= \displaystyle \lim_{n \to \infty} \left( \cos \dfrac{\pi}{\sqrt{n}} \right)^n \\ &= \displaystyle \exp \left( \lim_{n \to \infty} n \left( \cos \dfrac{\pi}{\sqrt{n}} - 1 \right) \right) \\ &= \displaystyle \exp \left( \lim_{n \to \infty} -2n \sin^2 \dfrac{\pi}{2 \sqrt{n}} \right) \\ &= \displaystyle \exp \left( \lim_{n \to \infty} -\dfrac{1}{2} \left( 2 \sqrt{n} \sin \dfrac{\pi}{2 \sqrt{n}} \right)^2 \right) \\ &= \displaystyle \exp \left( -\dfrac{ \pi^2}{2} \right) & \small\color{#3D99F6}{\text{ given } \displaystyle \lim_{n \to \infty} n \sin \dfrac{\pi}{n} = \pi }\\ &\approx 0.00719188 \end{aligned}$

Thus

$\left\lfloor 1000 A \right\rfloor = \boxed{7}$

Let 1/√(n) =x. Then n=1/(x^2). ln(A) is the limit of ln(cos(πx)) /(x^2) as x tends to zero. This is a 0/0 form. Applying L'Hospital's rule we get ln(A)=-(π^2)/2 or A=exp(-(π^2)/2)=0.00719188