Calculate $\lim \limits_{n \to \infty} (\sqrt{2n+1} - \sqrt{2n-1})$ using the sandwich theorem

So far I got:

$\sqrt{2n+1} > \sqrt{2n-1}$, so $\sqrt{2n+1} - \sqrt{2n-1} > 0$ for all $n > 0$.

Then I tried to find a sequencen, which converges to $0$ and is bigger than $\sqrt{2n+1} - \sqrt{2n-1}$. For this I tried to transform the expression:

\begin{align*} \sqrt{2n+1} - \sqrt{2n-1} &= \frac{(\sqrt{2n+1} - \sqrt{2n-1})(\sqrt{2n+1} - \sqrt{2n-1})}{\sqrt{2n+1} - \sqrt{2n-1}} \\ &= \frac{2n+1 - \sqrt{2n+1}\sqrt{2n-1} - \sqrt{2n-1}\sqrt{2n+1}+2n-1}{\sqrt{2n+1} - \sqrt{2n-1}} \\ &= \frac{4n}{\sqrt{2n+1} - \sqrt{2n-1}} \\ \end{align*}

Now I am having a hard time finding a proper sequence which is greater than $\frac{4n}{\sqrt{2n+1} - \sqrt{2n-1}}$ and converges to 0. Hence, I think I made some mistakes during my journey. Any hints?

The trick when you have this type of expression is to get rid of the square roots. You can use $$a^2-b^2=(a-b)(a+b)$$ In this case you have $a=\sqrt{2n+1}$ and $b=\sqrt{2n-1}$, so $$\sqrt{2n+1}-\sqrt{2n-1}=\frac{(\sqrt{2n+1})^2-(\sqrt{2n-1})^2}{\sqrt{2n+1}+\sqrt{2n-1}}=\frac2{\sqrt{2n+1}+\sqrt{2n-1}}$$ Can you take it from here?