# 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

## 1 Answer

You are right till there. Now divide both the numerator and denominator by 'n'. You get 4 / (1 + sqrt(1-4/n)) . as n tends to infinity, that gives 4/1+1 = 2;

• I am trying to use the definition of the limit and not assume that 4/n tends to 0 as n goes to infinity. – IntegrateThis Feb 7 '16 at 1:36
• @JamesDickens "I am trying to ... not assume that 4/n tends to 0" Wait, are you serious? – Did Feb 7 '16 at 7:14
• @Did I think he wants to do an epsilon-delta proof. – Akiva Weinberger Feb 7 '16 at 7:23
• @AkivaWeinberger ...Which, sooner or later, will need to prove or to admit that 1/n converges to 0. – Did Feb 7 '16 at 7:31
• @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