Yet another epsilon-delta proof

[bjgawp]

Homework Statement

Prove that \lim_{x \to -\infty} \frac{2}{\sqrt{x^{4}+1}}=0\

Homework Equations

The Attempt at a Solution

Preliminary Work
http://img339.imageshack.us/img339/401/proofzr7.jpg

Before proceeding on with the proof, when we look at the last line there seems to be a logical problem. We know that \epsilon > 0 and x < N < 0 . Thus, it seems counter-intuitive that x is greater than a positive expression but less than a negative. Just a guess but I think it has something to deal with the second last line involving the square root. Thanks in advance!

Edit: Are there any texts that you guys suggest for learning epsilon-delta proofs? The examples that we do in class seem to be repetitive but when it becomes more abstract and general, some ingenuinity is needed and I would like to see some of these proofs worked out. Thanks!

 

[ace123]

Should move this to calculus for better results and don't expect much. Delta-epsilon proof is a confusing topic

 

[Dick]

It sure is counterintuitive. How can you say that the solution set to |x|>C is C<x<-C? Don't you mean x<-C or x>C?? E.g. imagine C=2.

 

[bjgawp]

How do i get from the penultimate line to a solution for x? If I proceed:
x < \sqrt[4]{\frac{2}{\epsilon^{2}}-1} and set N to that when we give the formal proof, this would contradict one of our initial statements that N < 0 since the expression \sqrt[4]{\frac{2}{\epsilon^{2}}-1} must be positive. So I'm not sure what the solution set of x would be.

 

[klile82]

Can you use what you know about x<0? So from your next to last line, say
|x| &gt;\sqrt[4]{\frac{2}{\epsilon}-1}
-|x|&lt;\sqrt[4]{\frac{2}{\epsilon}-1}&lt; |x|, but you know that x=-|x|

Okay, I obviously need to work on my tex communication skills, but hopefully you can make sense of this.

 

[Dick]

Call C your fourth root expression. Yes, C is positive. You want |x|>C and x negative. You want x<-C. I don't see your 'contradictions'.

 

[bjgawp]

Ah it's all clicked in guys. So for the formal proof I'll set N = -C.

Edit: Sorry if I need to post another thread but just a quick question about a simpler proof.
Prove that \lim_{x \to -\infty} \frac{1}{x}=0

In the solutions manual:
For x < 0, |\frac {1}{x}-0|=- \frac{1}{x}. If \epsilon > 0 is given, then - \frac {1}{x}&lt; \epsilon which means x < - \frac{1}{\epsilon}.

Take N = - \frac{1}{\epsilon}. Then,

x &lt; N
x &lt; -\frac{1}{\epsilon}
|-\frac{1}{x} - 0| = -\frac{1}{x} &lt; \epsilon
So, \lim_{x \to -\infty} \frac{1}{x}=0

My problem is with the initial statement claiming that |\frac {1}{x}-0|=- \frac{1}{x}. I fail to see how this works as the absolute value of a function yields positive values, does it not? And wouldn't the proof work if we just made the condition that \frac{1}{x} &lt; 0 &lt; \epsilon , implying that x &lt; 0 &lt; \frac{1}{\epsilon}?

I apologize in advance if my train of thought isn't presented very clearly. Thanks again!

 

[klile82]

Again, your limit is as x approaches neg inf, so x is negative. So the absolute value |1/x -0| = -1/x is positive.

 

[klile82]

Also, since x<0, when you multiply both sides of 1/x < eps by x, the inequality flips. Does that make sense? You know that eps is positive and x is negative, so you can't say that 1/eps < x. Sorry for my lack of tex. I'm working on that.

 

[bjgawp]

Ah It all fell into place. Thanks a lot klile82 and Dick!

I'll definitely be back with more ... XD

 

[bjgawp]

Following Dick's advice and posting a new thread.

 

[Dick]

The statement you want to prove doesn't make much sense. Could you post it exactly as stated? And I would suggest you start another thread with a new problem. You'll get a lot more attention that way. If I see an active thread with over 5 posts, I don't generally don't even check it. Figure the issue is in the process of being beaten to death. Or that the poster is so confused no one can help.

 

[bjgawp]

Sure no problem. And i wrote the question wrong, sorry!