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I am attempting to decipher the word of Stewart but I can't really understand any of the epsilon delta stuff. I watched several videos online and I have a better understanding but I still don't quite get how to do the math as no one really describes that part. As far as I understand I need to find the distance between x and delta that is less than delta and greater than zero that will coorespond to an epsilon (y).

So I have the problem $$ \lim_{x \to 1}\frac{2+4x}{3}=2 $$ so I do some algebra magic and I get $x=1$ but from there I am not sure what to do.

Norbert
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2 Answers2

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The construction of an $\epsilon$-$\delta$ proof is usually exactly opposite its presentation. This should be a good example. Given $\epsilon$, you need to be able to produce some $\delta$ such that for all $x$ with $|x-1|<\delta$, $\frac{2 + 4x}{3} - 2|<\epsilon$.

How do you do this? You (usually) need some formula to produce $\delta$ in terms of $\epsilon$. So start with the expression you need, $$\bigg|\frac{2 + 4x}{3} - 2\bigg|<\epsilon,$$ and start solving for $x$ in terms of $\epsilon$.

I'll let you work out the details here; it's a simple algebraic exercise. At the end, you'll get something like $$\bigg|x - 1\bigg| < \frac{3}{4}\epsilon.$$

Ah-hah! Now you have produced a constraint on $|x-1|$ in terms of $\epsilon$. If I give you $\epsilon$, you can pick $\delta < \frac{3}{4}\epsilon$ and, as you can quickly verify, this $\delta$ will satisfy the $\epsilon$ bound. In fact, it has to --- that's how you made it.

The take-home lesson here, again, is that the way you build the proof is backwards. Start with what you want and backsolve for what you need. Then when you write the proof out, you know how to choose $\delta$, so you can quickly verify that $\lim_{x\to 1}\frac{2 + 4x}{3} = 2.$

Neal
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    So if I am given a number I just multiply it by 3/4 and that is my x value? –  Jan 22 '12 at 22:00
  • Right. Remember, the limit of a function at a point, if it exists, is the value the function approaches as its input approaches the point. So the limit exists at $p$ and is $L$ ($\lim_{x\to p}f(x) = L$) if you can make the function gets as close as you please ($|f(x) - L|<\epsilon$) by making $x$ very close to $p$ ($|x-p|<\delta$). Think of it as a game: I set how close you have to get (within $\epsilon$), and you have to be able to pick $\delta$ so that everything less than $\delta$ from $p$ gets sent closer than $\epsilon$ to $L$. – Neal Jan 22 '12 at 22:06
  • I don't get it, can't I just make up numbers? They have to work. Why don't I just make up a number .5 and plug that in to find the difference? –  Jan 22 '12 at 22:08
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    Because you have to be able to find a $\delta$ for EVERY $\epsilon$ that I give you. $0.5$ might work if I give you $2$, but if I give you $0.000001$, then $0.5$ won't work any more. – Neal Jan 22 '12 at 22:22
  • I don't understand this at all and Steward has 8 definitions and several proofs of epsilon delta and they all seem to either say nothing or they disagree with eachother. If you give me .000001 then can't I just add or subtract that from the limit and then plug that in for x? –  Jan 22 '12 at 22:30
  • Hmm. Are you trying to solve for $x$? – Neal Jan 22 '12 at 22:35
  • I am not sure, I am just trying to do the homework. "Prove the statement using the epsilon delta definition of a limit." –  Jan 22 '12 at 22:37
  • @Jordan - the definition of a limit, paraphrased a bit, is "No matter how small you make $\epsilon$, I'll be able to choose a small enough $\delta$ to make the inequality work out". The key point is that I don't have to choose $\delta$ first; I can wait till you've chosen $\epsilon$, then calculate a $\delta$ that depends on your $\epsilon$. So I can't just pick $\delta = 0.0001$ and hope for the best, because you could choose an $\epsilon$ that's really small, and that makes the inequality false. I need to find a way of expressing $\delta$ in terms of $\epsilon$, to make the inequality work. –  Jan 22 '12 at 22:55
  • Might I add, Jordan, that the definition of the limit is a logical implication. Think of it as P -> Q (P implies Q). In this case, the statement P is that "x is close enough to delta" and the Q is "f(x) is close enough to epsilon"... more or less. You might not know that to prove P -> Q, you assume P and deduce Q – The Chaz 2.0 Jan 22 '12 at 23:20
  • So that means that you can chose the value of the function as close to the limit as you want and that there should be an x value with a difference to the limit that is positive but smaller than the function change value? –  Jan 22 '12 at 23:37
  • @Jordan - Kind of, but by choosing $\epsilon$, you're actually choosing a whole range of values close to the limit (basically from $f(x) - \epsilon$ to $f(x) + \epsilon$; and saying that a whole range of $x$ values will map into this range. –  Jan 22 '12 at 23:54
  • Is epsilon larger or smaller than delta? –  Jan 23 '12 at 00:08
  • @Jordan - could be either, it depends on the question. In this particular case, $\delta$ has to be smaller. –  Jan 23 '12 at 00:29
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In order to calculate the limit using $\epsilon-\delta$ we have to prove that:

$ \lim_{x \to 1} \frac{2+4x}{3}=2 \Longleftrightarrow ( \forall \epsilon >0) (\exists \delta >0) ( \forall x) |x-1|< \delta \Longrightarrow |\frac{2+4x}{3}-2|<\epsilon$

So, choosing $ \delta= \frac{3}{4}\epsilon$ we get:

$ |\frac{2+4x}{3}-2|=\frac{4|x-1|}{3} <\frac{4 \delta}{3} =\epsilon$.

passenger
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  • What is the upside down A, what is the backwards E and why do you have arrows points both ways and some only one way? Why are you multiplying upsidedown A with things that are larger than 0? –  Jan 22 '12 at 21:55
  • These are shorthand for logical statements. $\forall$ = "for every," $\exists$ = "there exists some ... such that," $\Rightarrow$ = "implies," $\Leftrightarrow$ = "if and only if." So, for example, $\forall\epsilon > 0 \exists\delta > 0 \forall x |x-1|<\delta \Rightarrow |\frac{2+4x}{3}-2|<\epsilon$ translates to "for each $\epsilon > 0$ there is some $\delta > 0$ such that for all $x$, if $x$ is less than $\delta$ away from $1$, $\frac{2 + 4x}{3}-2$ is less than $\epsilon$ away from $2$. – Neal Jan 22 '12 at 21:58
  • @Jordan: $\forall= \text{for all}, \quad \exists=\text{exist}$ – passenger Jan 22 '12 at 21:59
  • The equivalence you wrote after we have to prove that is a tautology, hence it cannot be what one has to prove. – Did Jan 23 '12 at 00:30
  • The sentence with "we have to prove that" ends just before the double arrow. –  Jan 23 '12 at 00:41
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    @David: I hope you do not recommend such a writing, especially in the answer to a question of this level. – Did Jan 23 '12 at 00:53
  • No, I absolutely don't. At the very least, I would have written some words on a separate line, in place of the double arrow. Also, since it's supposed to be an explanation, not a proof, I would have put some words explaining how to get to $\delta = \frac{3}{4}\epsilon$. –  Jan 23 '12 at 01:04
  • @David: Then we agree. – Did Jan 23 '12 at 02:00