calculus - Find the limit of the sequence $left( sqrt {2n^{2}+n}-sqrt {2n^{2}+2n}right) _{nin N}$

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 $\lim _{n\rightarrow \infty }\left( \sqrt {2n^{2}+n}-\sqrt {2n^{2}+2n}\right) =0$

Answer

You may write, as $n \to \infty$,

$$\begin{align} \sqrt {2n^{2}+n}-\sqrt {2n^{2}+2n}&=\frac{(2n^{2}+n)-(2n^{2}+2n)}{\sqrt {2n^{2}+n}+\sqrt {2n^{2}+2n}} \\\\&=\frac{-n}{\sqrt {2n^{2}+n}+\sqrt {2n^{2}+2n}} \\\\&=\frac{-1}{\sqrt {2+1/n}+\sqrt {2+2/n}} \\\\&\to-\frac{1}{2\sqrt{2}} \end{align}$$