<s>[INST] $\lim_{n \to \infty} (\sqrt{9n^2 + 2n + 1} - 3n) = \frac{1}{3}$ Prove
$$\lim_{n \to \infty} (\sqrt{9n^2 + 2n + 1} - 3n) = \frac{1}{3}$$
I got this problem in Harro Heuser's "Lehrbuch der Analysis Teil 1". It is surely smaller than 1 because $\sqrt{9n^2 + 2n + 1} < \sqrt{9n^2 + 6n + 1} = 3n + 1$, but I cannot get closer than that, although it looks very simple...
 [/INST] $$\lim_{n\to\infty} (\sqrt{9n^2 + 2n+1} - 3n) = \lim_{n\to\infty} 3n\left(\sqrt{1 + \frac{2}{9n} + \frac{1}{9n^2}} - 1\right)$$
Now, using Taylor expansion for $\sqrt{1 + x} = 1 + \frac{x}{2} + o(x)$
$$\lim_{n\to\infty}3n\left(1 + \frac{1}{9n} + o\left(\frac{1}{n}\right) - 1\right) = \lim_{n\to\infty}\frac{3n}{9n} = \frac{1}{3}$$
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