\[\begin{split}&\lim_{n \to \infty} n\left(\sqrt{2n^2+1}-\sqrt{2n^2-1}\right)=\\[16pt] &=\lim_{n \to \infty} \frac{n\left(2n^2+1-(2n^2-1)\right)}{\left(\sqrt{2n^2+1}+\sqrt{2n^2-1}\right)}=\\[16pt] &=\lim_{n \to \infty} \frac{2n}{\left(\sqrt{2n^2+1}+\sqrt{2n^2-1}\right)}\frac{:n}{:n}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\dfrac{2n}{n}}{\sqrt{\dfrac{2n^2}{n^2}+\dfrac{1}{n^2}}+\sqrt{\dfrac{2n^2}{n^2}-\dfrac{1}{n^2}}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{2}{\sqrt{2+\dfrac{1}{n^2}}+\sqrt{2-\dfrac{1}{n^2}}}=\\[16pt] &=\frac{2}{2\sqrt{2}}=\frac{\sqrt{2}}{2}\end{split}\]