Simplify _Elementary Calculus_ section 1.6 problem 33

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.

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Divide top and bottom by $\sqrt{H}$. That does not change the value of the expression.

On top we get $\sqrt{1+\epsilon}$, where $\epsilon=1/H$.

On the bottom, we get $\sqrt{2}+\sqrt{1-\epsilon}$. If $H$ is infinitely large, then $\epsilon$ is infinitesimal. (Note that $H$ cannot be infinitely large negative, since if it were then our square roots would not be defined.)

Now finding the standard part is easy. The answer is $\dfrac{1}{\sqrt{2}+1}$. If you prefer, you can write this as $\sqrt{2}-1$.

Remark: The same idea is useful in many other cases. For example, if we are interested in the behaviour of $$\frac{3x^3+100x^2+700}{7x^3-77x^2+x}$$ when $x$ has large absolute value, divide top and bottom by $x^3$.

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$N$? You mean $H$? – rotten Feb 2 '12 at 6:05
@rotten: Yes, I mean $H$. The name $N$ sounded like a good one for an infinitely big number, I didn't notice it was called $H$ in the post. Fixed. – André Nicolas Feb 2 '12 at 6:09
Yeah, what gives? $H$ always goes to zero and $N$ always goes to infinity ;) – rotten Feb 2 '12 at 6:23
@rotten: Not quite, it is $h$ (little aitch) that often goes to $0$. – André Nicolas Feb 2 '12 at 6:31
Thanks again for your help! I am wondering though, is there a way to algebraically convert $\frac{1}{\sqrt{2}+1}$ into $\sqrt{2}-1$? – Daniel Lyons Feb 2 '12 at 20:28