# Determining hyperreal class for $\frac{\epsilon + \delta}{\sqrt{\epsilon^2 + \delta^2}}$

I'm solidifying my calculus by going through Keisler's book that uses a hyperreal/infinitesimal approach. I'm stuck on this problem.

Given infinitesimals $\epsilon,\delta > 0$, deterimine whether the following expression is infinitesimal, finite but not infinitesimal, or infinite:

$$\frac{\epsilon + \delta}{\sqrt{\epsilon^2 + \delta^2}}$$

Keisler gives this hint: Assume $\epsilon \geq \delta$ and divide through by $\epsilon$.

Being an odd number problem, the answer is in the back, but I don't understand how to get it. Also, why would I assume that $\epsilon \geq \delta$?

• You can assume that because it is symmetric. One is necessarily bigger, so might as well be $\epsilon$ – Valtteri Mar 1 '13 at 15:26
• Thanks, I suspected that I could assume $\delta$ was larger just as well, so thanks for explaining that. – labyrinth Mar 1 '13 at 15:32
• To elaborate: It is a nice shortcut if you assume $\epsilon \geq \delta$ and prove the result, you just "rename" $\epsilon$ and $\delta$ in the proof to get the case where $\delta > \epsilon$. This is a common "trick," often accompanied with "Without loss of generality(WLOG), assume..." where some sort of symmetry is evident (as Valtteri points out) – Tyler Mar 1 '13 at 15:32
• Thanks, Tavares, for fixing my title. This is my first post here and didn't realize I had to use the dollar signs in the title, too. – labyrinth Mar 1 '13 at 15:34

Well, my take. If $\epsilon \ge \delta$, then $\frac{\epsilon+\delta}{\sqrt{\epsilon^2+\delta^2}} = \frac{1+\frac{\delta}{\epsilon}}{\sqrt{1+\delta^2/\epsilon^2}}$. Now $\delta \le \epsilon,$ so $\delta/\epsilon \le 1$ (because the resulting sequence will have smaller numbers divided by larger numbers a number of times that is in the ultrafilter).
• At first I thought that it amounted to 1, but now I see that in the case where $\epsilon = \delta$, it could be $\frac{2}{\sqrt{2}} = \sqrt{2}$ – labyrinth Mar 1 '13 at 15:58