Odpowiedź:
[tex]\lim\limits_{n\to\infty}\left(\sqrt{2n^2-4n+7}-\sqrt{2}n\right)=-\sqrt{2}[/tex]
Szczegółowe wyjaśnienie:
[tex]\lim\limits_{n\to\infty}\left(\sqrt{2n^2-4n+7}-\sqrt{2}n\right)=\\\\\lim\limits_{n\to\infty}\left(\sqrt{2n^2-4n+7}-\sqrt{2n^2}\right)=\\\\\lim\limits_{n\to\infty}\left(\left(\sqrt{2n^2-4n+7}-\sqrt{2n^2}\right)\cdot\dfrac{\sqrt{2n^2-4n+7}+\sqrt{2n^2}}{\sqrt{2n^2-4n+7}+\sqrt{2n^2}}\right)=\\\\\lim\limits_{n\to\infty}\left(\dfrac{2n^2-4n+7-2n^2}{\sqrt{2n^2-4n+7}+\sqrt{2n^2}}\right)=\\\\\lim\limits_{n\to\infty}\left(\dfrac{-4n+7}{\sqrt{2n^2-4n+7}+\sqrt{2n^2}}\right)=[/tex]
[tex]\lim\limits_{n\to\infty}\left(\dfrac{-4+\frac{7}{n}}{\sqrt{2-\frac{4}{n}+\frac{7}{n^2}}+\sqrt{2}}\right)=\dfrac{-4+0}{\sqrt{2-0+0}+\sqrt{2}}=\dfrac{-4}{2\sqrt{2}}=\boxed{-\sqrt{2}}[/tex]
