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The approximate value of the expression $$2\log_3 3n - \log_3(n^2 + 1)$$ for a sufficiently large $$n$$ is
$$2\log_3(3n) - \log_3(n^2 + 1) = \log_3\dfrac{9n^2}{n^2 + 1}$$.
As $$n \to \infty$$, $$\dfrac{9n^2}{n^2+1} \to 9$$. So the expression $$\to \log_3 9 = \mathbf{2}$$.
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