# proof of a limit as n→∞

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• May 21st 2009, 03:56 PM
thisguy
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 =)
• May 21st 2009, 04:07 PM
Krizalid
by rationalizing you get that $\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 $\sqrt n$ to get the limit.
• May 21st 2009, 04:24 PM
thisguy
thanks, would you mind explaining how you rationalized this please?
• May 21st 2009, 04:26 PM
Spec
$\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})}}=
$
$\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n+1}}$

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