The graph of \(f(x) = sin(\frac 1 x)\) gets a little...crazy near \(x=0\). Specifically, the graph of \(f(x)\) has an infinite number of zero-crossings (roots, or values of \(x\) such that \(f(x) = 0\)) between the first one at \(x=\frac{1}{\pi}\) and \(x=0\).
###### Image credit: Wolfram Alpha

Let \(x_n\) be the \(n^{th}\) zero-crossing of the function \(f(x)\), where \(x_1 = \dfrac{1}{\pi}\), and \(x_2 = \dfrac{1}{2\pi}\) and so on in the negative direction. Also let \(f'(x)=\dfrac{df(x)}{dx}\).

Find \(\displaystyle \lim_{n \to \infty} \frac{f'(x_{n+1})}{f'(x_n)}\)

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