# Epsilon Delta and The Triangle Inequality

arildno
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Now, the first thing you should do, is to ascertain that what I have done is VALID.

Thank you for sticking around and helping me.

I have gone through the steps, and I see that how it is all valid.

I do have a few particulars I still dont understand, as you predicted.

1)you choose $$\delta < \delta_1 = 1$$ arbitrarily.. Is that enable you to solve the delta expression in terms of epsilon?

2)When you did this $$\delta^2 + 6x < \delta\delta_1 + 6\delta = 7\delta$$ why did you choose to replace only one of the deltas with 1? And why did you choose to replace that particular delta?

3)How do we know this is all true if $$[|x^2 - 9|<\epsilon$$? Simply because it was stated in the begining?

4)The maxium values $$\delta$$ can be are $$1$$or$$\epsilon/7$$ right? I dont understand why one has epsilon in it and one does not. Why is this and what does it imply?

Thx again for helping me through this. Our teacher just glanced over it and didnt even test us on it, but I still really want to know. It is late and I am tired so I hope this made sense.

Pyrrhus
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What part didnt I read as it was presented?

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arildno
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Gold Member
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DieCommie said:
Thank you for sticking around and helping me.

I have gone through the steps, and I see that how it is all valid.
I do have a few particulars I still dont understand, as you predicted.
1)you choose $$\delta < \delta_1 = 1$$ arbitrarily.. Is that enable you to solve the delta expression in terms of epsilon?
I did it because it would simplify my work further on; it certainly wasn't a NECESSARY step.
2)When you did this $$\delta^2 + 6x < \delta\delta_1 + 6\delta = 7\delta$$ why did you choose to replace only one of the deltas with 1? And why did you choose to replace that particular delta?
Again, a VALID simplification trick.[/QUOTE]
3)How do we know this is all true if $$[|x^2 - 9|<\epsilon$$? Simply because it was stated in the begining?
OOOPS!
Here you've misunderstood the issue!
If we CHOOSE $$\delta$$ to fulfill $$\delta<min(1,\frac{\epsilon}{7})$$, then $$|x-3|<\delta$$ IMPLIES that $$|x^{2}-9|<\epsilon$$
This is what we're after, not the other way around!!!
The "restrictions" (or conditions) I mentioned are to be understood in the following manner:
"What restrictions/conditions must be placed upon $$\delta$$ so that $$|x-3|<\delta$$ IMPLIES $$|x^{2}-9|<\epsilon$$?"
4)The maxium values $$\delta$$ can be are $$1$$or$$\epsilon/7$$ right? I dont understand why one has epsilon in it and one does not. Why is this and what does it imply?
Read again, in particular my comment on 3), and if you still have problems, let's tackle those.

Remember that to be able to make valid and smart simplification tricks is a matter of experience (you'll develop a skill in this through practice).

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