Originally Posted by

**Krizalid** I'll take it from here:

$\displaystyle \frac{{\left( {\sqrt {n + 2\sqrt n } - \sqrt n } \right)\left( {\sqrt {n + 2\sqrt n } + \sqrt n } \right)}}

{{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{{n + 2\sqrt n - n}}

{{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{{2\sqrt n }}

{{\sqrt {n + 2\sqrt n } + \sqrt n }}.$

Now, divide by $\displaystyle \sqrt n$ top & bottom, so

$\displaystyle \frac{{2\sqrt n }}

{{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{2}

{{\sqrt {1 + \dfrac{2}

{{\sqrt n }}} + 1}}.$

The conclusion follows.

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By the way, applyin' L'Hôpital's Rule, it's a mess, with all those roots.