👤

Va rog sa ma ajutati! Limite clasa a 11a - rezolvate simplu, fara l'Hopital. Multumesc!!

[tex]\lim_{n \to \infty} \sqrt{n^2+1}-n\sqrt{n}[/tex]


Răspuns :

[tex] n\to \infty \Rightarrow \sqrt{n^2+1} \approx \sqrt{n^2} [/tex]

Astfel:

[tex]\lim\limits_{n\to\infty} \left(\sqrt{n^2+1} - n\sqrt{n}\right) = \lim\limits_{n\to\infty} \left(\sqrt{n^2} - n\sqrt{n}\right) =[/tex]

[tex] = \lim\limits_{n\to\infty} \left(|n| - n\sqrt{n}\right) = \lim\limits_{n\to\infty} \left(n - n\sqrt{n}\right) = \lim\limits_{n\to\infty} n\left(1-\sqrt{n}\right) =[/tex]

[tex] = \infty\cdot (-\infty) = -\infty[/tex]