# Thread: proof of a limit as n→∞

1. ## proof of a limit as n→∞

A friend showed me this and neither of us can work out a proof.

lim n→∞((sqroot(n+2)-sqroot(n+1))/(sqroot(n+1)-sqroot(n)))

I know the limit is 1 but I'm just looking for a method to prove it. So not putting in high values and seeing what it tends to and doing it by inspection probably wouldn't be ideal. I've tried multiplying the numerator and denominator by several values such as sqroot(n+1)/sqroot(n+1) and hit dead ends with all of them so I'm assuming there's a different approach I'm not seeing. Thanks for any help in advance =)

2. by rationalizing you get that $\displaystyle \frac{\sqrt{n+2}-\sqrt{n+1}}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+2}+\s qrt{n+1}},$ then just divide top & bottom by $\displaystyle \sqrt n$ to get the limit.

3. thanks, would you mind explaining how you rationalized this please?

4. $\displaystyle \frac{\sqrt{n+2}-\sqrt{n+1}}{\sqrt{n+1}-\sqrt{n}}=\frac{{\color{blue}(\sqrt{n+2}+\sqrt{n+1 })}(\sqrt{n+2}-\sqrt{n+1}){\color{red}(\sqrt{n+1}+\sqrt{n})}}{{\c olor{blue}(\sqrt{n+2}+\sqrt{n+1})}(\sqrt{n+1}-\sqrt{n}){\color{red}(\sqrt{n+1}+\sqrt{n})}}=$ $\displaystyle \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n+1}}$

EDIT: Trying to figure out how colors work in Latex so I can make it more clear.