<s>[INST] Find the limit of the sequence $\left( \sqrt {2n^{2}+n}-\sqrt {2n^{2}+2n}\right) _{n\in N}$ My answer is as follows, but I'm not sure with this:
$\lim _{n\rightarrow \infty }\dfrac {\sqrt {2n^{2}+n}}{\sqrt {2n^{2}+2n}}=\lim _{n\rightarrow \infty }\left( \dfrac {2n^{2}+n}{2n^{2}+2n}\right) ^{\dfrac {1}{2}}$
$\lim _{n\rightarrow \infty }\dfrac {2n^{2}+n}{2n^{2}+2n}=\lim _{n\rightarrow \infty }\dfrac {2+\dfrac {1}{n}}{2+\dfrac {2}{n}}$
since $\lim _{n\rightarrow \infty }\dfrac {1}{n}=0$, $\lim _{n\rightarrow \infty }\dfrac {2n^{2}+n}{2n^{2}+2n}=1$
hence $\lim _{n\rightarrow \infty }\left( \dfrac {2n^{2}+n}{2n^{2}+2n}\right) ^{\dfrac {1}{2}}=\left( 1\right) ^{\dfrac {1}{2}}=1$ (by composite rule)
hence $\sqrt {2n^{2}+n}=\sqrt {2n^{2}+2n}$ as $n\rightarrow \infty $
so $\lim _{n\rightarrow \infty }\left( \sqrt {2n^{2}+n}-\sqrt {2n^{2}+2n}\right) =0$
 [/INST] Multiplying it by
$$\frac{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}$$
gives
$$\begin{align}&\lim_{n\to\infty}(\sqrt{2n^2+n}-\sqrt{2n^2+2n})\\\\&=\lim_{n\to\infty}(\sqrt{2n^2+n}-\sqrt{2n^2+2n})\cdot\frac{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}\\\\&=\lim_{n\to\infty}\frac{(2n^2+n)-(2n^2+2n)}{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}\\\\&=\lim_{n\to\infty}\frac{-n}{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}\\\\&=\lim_{n\to\infty}\frac{-1}{\sqrt{2+\frac 1n}+\sqrt{2+\frac{2}{n}}}\\\\&=-\frac{1}{2\sqrt 2}\end{align}$$
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