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 $\lim _{n\rightarrow \infty }\left( \sqrt {2n^{2}+n}-\sqrt {2n^{2}+2n}\right) =0$
 A: You may write, as $n \to \infty$,
$$
\begin{align}
\sqrt {2n^{2}+n}-\sqrt {2n^{2}+2n}&=\frac{(2n^{2}+n)-(2n^{2}+2n)}{\sqrt {2n^{2}+n}+\sqrt {2n^{2}+2n}}
\\\\&=\frac{-n}{\sqrt {2n^{2}+n}+\sqrt {2n^{2}+2n}}
\\\\&=\frac{-1}{\sqrt {2+1/n}+\sqrt {2+2/n}}
\\\\&\to-\frac{1}{2\sqrt{2}}
\end{align}
$$
A: Multiplying it by
$$\frac{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}$$
gives
$$\begin{align}&\lim_{n\to\infty}(\sqrt{2n^2+n}-\sqrt{2n^2+2n})\\\\&=\lim_{n\to\infty}(\sqrt{2n^2+n}-\sqrt{2n^2+2n})\cdot\frac{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}\\\\&=\lim_{n\to\infty}\frac{(2n^2+n)-(2n^2+2n)}{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}\\\\&=\lim_{n\to\infty}\frac{-n}{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}\\\\&=\lim_{n\to\infty}\frac{-1}{\sqrt{2+\frac 1n}+\sqrt{2+\frac{2}{n}}}\\\\&=-\frac{1}{2\sqrt 2}\end{align}$$
A: If you know that for small $x$ we have that $\sqrt{1+x}$ behaves like $1+\frac{x}{2}$, then $\sqrt{2n^2+2an} = \sqrt{2}\cdot\sqrt{(n+a)^2-a^2}$ behaves like $\sqrt{2}(n+a)\left(1-\frac{a^2}{2(n+a)^2}\right)$, or $n\sqrt{2}+a\sqrt{2}+O\left(\frac{1}{n}\right)$, for large $n$. To compute the limit from that is straightforward.
