$\sqrt{11}$ as rational number within $10^{-4}$?

Use continued fractions to find a rational number which approximates $\sqrt{11}$ to within $10^{−4}$.

I know how to solve for continued fractions like this: $$\sqrt{11}=3+x$$ $$11=9+6x+x^2$$ $$11=9+(6+x)x$$ $$2=x(6+x)$$ $$x=\frac{2}{6+x}=\frac{1}{3+\frac{x}2}=\frac{1}{3+\frac{1}{6+x}}$$

therefore, $$\sqrt{11}=[3;\overline{3,6}]$$

How do I find out where to terminate the fraction so as to obtain close value with error less than the mentioned limit

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