# Limit using definition

I am trying to find

$$\lim\limits_{x \to \infty} n-\sqrt{n^2-4n}$$ using the definition of a limit.

I have tried to multiply top and bottom by $n+\sqrt{n^2-4n}$ giving $\frac{4n}{n+\sqrt{n^2-4n}}$ but this does not seem to help too much. Any hints would be much appreciated. Thank you.

• Hint: $$0\leqslant\left(n-\sqrt{n^2-4n}\right)-2=\frac4{n-2+\sqrt{n^2-4n}}\leqslant{}{}{}{}\frac{4}{n-2}\leqslant\frac8{n}\quad (n\geqslant4)$$ But this approach will require to be able to prove that $8/n\to0$... – Did Feb 7 '16 at 7:34

• @Did Yeah. He doesn't want to assume it tends to zero, he wants to prove it tends to zero. (And he doesn't want to assume that $\lim f/g=\lim f/\lim g$ either, I bet.) – Akiva Weinberger Feb 7 '16 at 15:39