# Homework Help: Trying to fully grasp limit definition

1. Nov 15, 2008

### snipez90

1. The problem statement, all variables and given/known data
An example in the text that involves showing that $$x^2sin\frac{1}{x}$$ approaches 0 as x approaches 0.

2. Relevant equations
$$\epsilon -\delta$$ argument

3. The attempt at a solution
I can prove many limits efficiently now using $$\epsilon -\delta$$ but I don't think I am that flexible with the definition. I don't feel that I fully understand it. For instance, in this example, it's easy to choose $$d = \sqrt{\epsilon}$$ and after noting that $$|sin\frac{1}{x}| \leq 1$$, the proof is very short.

But in this part of the text, Spivak assumes that the reader does not know the definition yet. He argues that for $$|x^2sin\frac{1}{x}|$$ to be less than $$\epsilon$$, it is only required that $$|x| < \epsilon$$ and $$x \neq 0$$, provided that $$\epsilon \leq 1$$. This makes sense because $$|x^2| = |x|^2 \leq |x|$$ for $$|x| \leq 1$$ and hence the stated bound $$\epsilon \leq 1$$. If $$\epsilon > 1$$, then it is required that $$|x| < 1$$ and $$x \neq 0$$.

This approach may seem more trouble than it's worth since $$\delta = \sqrt{\epsilon}$$ apparently works well. But when trying to write up a proof based on the above approach, I had a hard time. I understood his approach but it seemed weird to be considering two different epsilon cases. After working off of $$|f(x) - L|$$ , I quickly got to $$|x^2sin\frac{1}{x}| \leq |x|^2$$, but am now stuck. I know from Spivak's argument that I can have $$|f(x)-L| < |x|^2 < |x| < \epsilon$$ for $$\epsilon \leq 1$$ but how do I choose $$\delta$$? Does choosing delta = min{a,b}, where a and b are real numbers, have something to do with this approach?

I guess choosing delta to be the min of two real quantities is still unclear to me. Basically for harder limit proofs, I manipulate the |f(x) - L| term until I get a quantity that has the |x-a| term (which is < delta). Then if multiple terms are involved, I assume that delta is bounded above by some z > 0 (usually 1 or a fraction less than 1) and find a bound for each of the terms beside the |x-a| term. Multiplying the bounds on these terms gives a bound B on the |x-a| term so that $$B|x-a| < B\delta$$. Then I just choose delta to be $$\frac{\epsilon}{B}$$.

Right now, my idea of why this approach is justified is that we can choose delta. Once we do enough manipulations to find the other delta choice dependent on epsilon, then epsilon can vary so that we can always find a delta for which the limit definition holds. But this is all very hazy and if someone could clarify my reasoning or justify my approach and why it works it would be greatly appreciated.

2. Nov 15, 2008

### Office_Shredder

Staff Emeritus
That's exactly it. If you can find an easy proof of the limit for small epsilon, say epsilon smaller than M, then when epsilon is larger than M, you can pick a delta that worked when epsilon was smaller than M and it still trivially works

3. Nov 15, 2008

### snipez90

Hmm ok I think that makes sense. For some reason I kept thinking that min meant that I could only pick one of the quantities that delta can equal. But since delta bounds two quantities, then no matter what epsilon > 0 is chosen, then the |f(x) - L| term will be less than epsilon.