Once again, I'm trying to simplify an expression from Elementary Calculus with hyperreals. Given that $H$ is infinite, compute the standard part of:
$\frac{\sqrt{H+1}}{\sqrt{2H}+\sqrt{H-1}}$
The answer in the back of the book is $\frac{1}{1+\sqrt{2}}$ or $\sqrt{2}-1$, which is consistent with what my calculator gives me if I compute: $\lim_{x\to\infty} \left(\frac{\sqrt{x+1}}{\sqrt{2x}+\sqrt{x-1}}\right)$
I see two ways to proceed: multiply the top and bottom by $\sqrt{H+1}$ to eliminate the radical in the numerator or multiply by the conjugate $\sqrt{2H}-\sqrt{H-1}$ but neither one seems to improve the situation.
Idea 1 (multiply by numerator):
$\frac{\sqrt{H+1}\sqrt{H+1}}{\sqrt{H+1}(\sqrt{2H}+\sqrt{H-1})}$ $=\frac{H+1}{\sqrt{H+1}(\sqrt{2H}+\sqrt{H-1})}$ $=\frac{H+1}{\sqrt{H+1}\sqrt{2H}+\sqrt{H+1}{\sqrt{H-1}}}$
Idea 2 (multiply by conjugate of denominator):
$\frac{\sqrt{H+1}(\sqrt{2H}-\sqrt{H-1})}{(\sqrt{2H}+\sqrt{H+1})(\sqrt{2H}-\sqrt{H+1})}$ $=\frac{\sqrt{H+1}(\sqrt{2H}-\sqrt{H-1})}{2H-(H+1)}$ $=\frac{\sqrt{H+1}(\sqrt{2H}-\sqrt{H-1})}{H-1}$
I don't think either of these are on the right path, but I don't know any other possibilities. Eventually, I suspect I will be multiplying the numerator and denominator by $\frac{1}{H}$ to convert some of these infinities to infinitesimals but I don't see a way to reduce it further (if these could be called reductions) before taking standard parts, and for some reason I feel like the algebraic manipulations should be complete before the standard part/hyperreal logic is used to eliminate infinitesimals.
Thanks for your help!