# Question of Rigour

1. Aug 31, 2009

### JG89

I was trying to prove a limit, and I found that I can take $$\epsilon = \sqrt[]{13}(\sqrt[]{\delta + 1} - 1)$$ since as delta goes to 0, it's obvious that entire expression also goes to 0, so epsilon also goes to 0.

My question is, how is the rigorousness of this? Will it be good enough for a uni. prof? (I have no experience with them)

2. Aug 31, 2009

### arildno

Actually, you should end up with choosing the delta, letting the epsilon be the arbitrary quantity.

Your profs will put their claws into that mistake of yours!.

So, rearrange your estimate to choosing delta as:
$$\delta\leq(\frac{\epsilon}{\sqrt{13}}+1)^{2}-1$$

3. Aug 31, 2009

### JG89

Thanks. It makes more sense when I re-arrange for delta. I know the steps are clearly reversible, but I guess I should get into the habit of solving for delta.

4. Aug 31, 2009

### arildno

Indeed you should!

5. Aug 31, 2009

### Tac-Tics

I'd say it's pretty obvious. It depends on the class and teacher I'd think. For a first-year analysis class, take the steps to prove it. If you're doing differential geometry, it's too basic to bother with.

If you want to avoid stupid pokes at your grade, you can always summarize a proof:

"The expression above approaches zero. $$\sqrt[]{13}(\sqrt[]{\delta + 1} - 1)$$ approaches zero if $$\sqrt[]{\delta + 1} - 1$$ approaches zero, $$\sqrt[]{\delta + 1} - 1$$ approaches 0 if $$\sqrt[]{\delta + 1}$$ approaches one, and $$\sqrt[]{\delta + 1}$$ approaches one if $$\delta + 1$$ does, which it does.

Or more concisely, you can explain that since the $$\sqrt{13}$$ factor, and the other terms are constant functions, it simply follows from the algebraic rules for limits that the whole deal is zero in the limit.

6. Aug 31, 2009

### arildno

Well, since OP couldn't possibly be studying higher order topics like differential geometry and still be uncertain about error estimation, I chose to regard OP as a bright student just having learnt about the epsilo-delta formalism.

7. Aug 31, 2009

### snipez90

Wait, what the heck? What limit are you trying to prove and why is there so much talk about epsilons and deltas approaching zero? Isn't it obvious that in most cases, if delta is to depend on epsilon, which is just an arbitrary positive number, then the smaller the epsilon that you are given, your delta will usually get smaller as well?

I mean I'm assuming the only reason you found delta so explicitly is so that you can estimate |f(x) - L| by making it less than epsilon, in which case it suffices to just make sure delta is actually positive?

8. Sep 2, 2009

### zhentil

If they're asking you to find delta explicitly, you have to show every single step. They're testing to see if you understand the definitions, and forgetting which constant is arbitrary is a dead giveaway.

9. Sep 2, 2009

### g_edgar

Depends where in your uni. career this is. For example, if this is early, before the existence of square roots has been proved, you are not going to get away with this. Also "It's obvious" may not be a good enough reason in some contexts.

10. Sep 2, 2009

### JG89

I figured. I haven't actually started uni. yet, I start next week :)

I really don't want to lose marks over stupid mistakes. I will just take arildno's advice and always express delta explicitly.