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...
$$ \lim_{n\to\infty} (\sqrt{9n^2 + 2n+1} - 3n) = \lim_{n\to\infty} \frac{\sqrt{9n^2 + 2n+1}^2-(3n)^2}{\sqrt{9n^2 + 2n+1} + 3n} = \\= \lim_{n\to\infty} \frac{2n+1}{3n\underbrace{\left(1 + \sqrt{1 + \frac29n+\frac{1}{n^2}}\right)}_{\to 2}}=\frac13 $$
Notice, $$\lim_{n\to \infty}(\sqrt{9n^2+2n+1}-3n)$$ Let, $n=\frac{1}{t} \implies t\to 0\ as\ n\to \infty$ $$=\lim_{t\to 0}\left(\sqrt{9\left(\frac{1}{t}\right)^2+2\left(\frac{1}{t}\right)+1}-3\left(\frac{1}{t}\right)\right)$$ $$=\lim_{t\to 0}\frac{\sqrt{t^2+2t+9}-3}{t}$$ Now, applying L-hospital's rule for $\frac{0}{0}$ form $$=\lim_{t\to 0}\frac{\frac{d}{dt}(\sqrt{t^2+2t+9}-3)}{\frac{d(t)}{dt}}$$ $$=\lim_{t\to 0}\frac{\frac{(2t+2)}{2\sqrt{t^2+2t+9}}-0}{1}$$ $$=\lim_{t\to 0}\frac{2(t+1)}{2\sqrt{t^2+2t+9}}$$$$=\lim_{t\to 0}\frac{t+1}{\sqrt{t^2+2t+9}}$$ $$=\frac{0+1}{\sqrt{0+2(0)+9}}$$ $$=\frac{0+1}{3}=\color{blue}{\frac{1}{3}}$$
$$\lim_{n\to\infty} (\sqrt{9n^2 + 2n+1} - 3n) = \lim_{n\to\infty} 3n\left(\sqrt{1 + \frac{2}{9n} + \frac{1}{9n^2}} - 1\right)$$
Now, using Taylor expansion for $\sqrt{1 + x} = 1 + \frac{x}{2} + o(x)$
$$\lim_{n\to\infty}3n\left(1 + \frac{1}{9n} + o\left(\frac{1}{n}\right) - 1\right) = \lim_{n\to\infty}\frac{3n}{9n} = \frac{1}{3}$$
$$\sqrt{9n^2 + 2n + 1} - 3n = \frac{(\sqrt{9n^2 + 2n + 1} - 3n)(\sqrt{9n^2 + 2n + 1} + 3n)}{\sqrt{9n^2 + 2n + 1} + 3n} = \frac{2n + 1}{\sqrt{9n^2 + 2n + 1} + 3n}$$
– Matthew Cassell Jul 22 '15 at 14:33