<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] 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  $$=\lim_{t\to 0}\frac{\frac{d}{dt}(\sqrt{t^2+2t+9}-3)}{\frac{d(t)}{dt}}$$ $$=\lim_{t\to 0}\frac{\frac{(2t+2)}{2\sqrt{t^2+2t+9}}-0}{1}$$
$$=\lim_{t\to 0}\frac{2(t+1)}{2\sqrt{t^2+2t+9}}$$$$=\lim_{t\to 0}\frac{t+1}{\sqrt{t^2+2t+9}}$$ $$=\frac{0+1}{\sqrt{0+2(0)+9}}$$ $$=\frac{0+1}{3}=\color{blue}{\frac{1}{3}}$$ 
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