<s>[INST] How to prove $\lim\limits_{x \to \infty} \frac{\pi\sqrt{x}}{\sin{(\frac{\pi}{\sqrt{x}} })} - x = \frac{\pi^2}{6}$? As we know, the result of some expressions and series is equal to $\frac{\pi^2}{6} $ that the most important of them is $\zeta(2)$
Now, I have founded an equation whose limit at point $x=\infty$ is equal to $\frac{\pi^2}{6} : $
$$\lim_{x \to \infty}  \frac{\pi\sqrt{x}}{\sin{(\frac{\pi}{\sqrt{x}}  })} - x = \frac{\pi^2}{6}$$
I want to know how can we prove it?
 [/INST] For simplicity let $x = \frac{1}{u^2}$. Then the limit becomes,
$$\lim_{u \to 0} (\frac{\pi u - \sin(\pi u)}{u^2 \sin(\pi u)})$$
This has indeterminate form of $\frac{0}{0}$, so we can apply l'hospitals rule,
$$=\lim_{u \to 0} (\frac{\pi - \pi \cos(\pi u)}{2u \sin(\pi u) + \pi u^2 \cos(\pi u)}) = \pi \lim_{u \to 0} (\frac{1 - \cos(\pi u)}{2u \sin(\pi u) + \pi u^2 \cos(\pi u)})$$
This has indeterminate form of $\frac{0}{0}$, so we can apply l'hospitals rule again,
$$=\pi \lim_{u \to 0} (\frac{\pi \sin(\pi u)}{2 \sin(\pi u) + 4 \pi u \cos(\pi u) - \pi^2u^2\sin(\pi u)}) = \pi^2 \lim_{u \to 0} (\frac{1}{2 + 4 \pi u \cot(\pi u) - \pi^2u^2})$$
Setting $v = \pi u$ simplifies the limit to,
$$=\pi^2 \lim_{v \to 0} (\frac{1}{2 + 4 v \cot(v) - v^2})$$
Now using the Maclaurin expansion of $\cot(v)$,
$$\pi^2 \lim_{v \to 0} (\frac{1}{2 + 4v(\frac{1}{v}-\frac{v}{3}-\frac{v^3}{45} - ...) - v^2}) = \pi^2 \lim_{v \to 0} (\frac{1}{2 + 4-\frac{4v^2}{3}-\frac{4v^4}{45} - ... - v^2})$$
Now setting $v=0$ the limit reduces to,
$$=\pi^2 (\frac{1}{2 + 4 - 0 - 0 - ... - 0}) = \frac{\pi^2}{6}$$
As required.
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