# Epsilon delta proof of a limit problems

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.

• Could you describe the magic? The limit is at $x = 1$, so that seems okay. – Dylan Moreland Jan 22 '12 at 21:44
• Oh I just meant I multiplied by 3, subtracted 2 and then divided by 4. – toby yeats Jan 22 '12 at 21:46
• I'll try to write something later, but here's a search for related questions. There are worked problems in a lot of these threads and that might help you. – Dylan Moreland Jan 22 '12 at 21:48

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 $\lim_{x\to 1}\frac{2 + 4x}{3} = 2.$

• So if I am given a number I just multiply it by 3/4 and that is my x value? – toby yeats 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? – toby yeats Jan 22 '12 at 22:08
• 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? – toby yeats Jan 22 '12 at 22:30

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:

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

• 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? – toby yeats 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 0:30
• The sentence with "we have to prove that" ends just before the double arrow. – user22805 Jan 23 '12 at 0:41