# Thread: Limits: Epsilon & Delta

1. ## Limits: Epsilon & Delta

Hey everyone. I have a question on the definition of limits basically. First, the definition states:
----------------------------------------
Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then we say taht the limit of f(x) as x approaches a is L, and we write
$\lim_{x\rightarrow a} f(x) = L$

if for every number $\epsilon > 0$ there is a corresponding number $\delta > 0$ such that

if $0 < |x - a| < \delta$ then $|f(x) - L| < \epsilon$
-------------------------
So the definition is basically saying that if the difference between x -a (whether x is approaching from the left/right since its abs value bar) is approaching 0, then the difference between f(x) - L is approaching 0. Is that correct?

Now an example:

Prove that $\lim_{x\rightarrow3} (4x - 5) = 7$

Solution:

So we say: if
$0 < |x - 3| < \delta$ then $|(4x - 5) - 7| < \epsilon$

$|4x - 12| < \epsilon$

$4|x - 3| < \epsilon$

$0 < |x - 3| < \frac{\epsilon}{4}$

$0 < |x - 3| < \delta$

And that's the end of it. I don't see how that proves anything besides that delta and epsilon are proportional. How does this prove anything?

Another question: What is the point of epsilon/delta? Can't you just prove this using right/left hand limits?

2. Hey there,

so the idea is that given any epsilon, you can select a delta that will satisfy the inequality.

the tricky part about these proofs are that you are essentially working backwards.

So, in this example, you "start" by saying,

let $\delta = \frac{\epsilon}{4}$

Then we have to prove that
$0 < |x - 3| < \delta$ implies $|(4x - 5) - 7| < \epsilon$
so to prove P implies Q, start by assuming P and show that Q holds.
so assume that
$0 < |x - 3| < \delta$
then
$0 < |x - 3| < \frac{\epsilon}{4}$
since $\delta = \frac{\epsilon}{4}$
(we got to "pick" $\delta$)
then
$4|x - 3| < \epsilon$
then
$|4x - 12| < \epsilon$
then
$|(4x - 5) - 7| < \epsilon$

so since,
$0 < |x - 3| < \delta$
Implies that
$|(4x - 5) - 7| < \epsilon$
then we're done.

so by the definition of limit,

for each $\epsilon > 0$ there is a $\delta > 0$ (in fact, $\delta = \frac{\epsilon}{4}$)
such that
$0 < |x - 3| < \delta$
Implies that
$|(4x - 5) - 7| < \epsilon$
then you can say
$
\lim_{x\rightarrow3} (4x - 5) = 7$

Hope that may make things clearer....

And also....
What is the point of epsilon/delta? Can't you just prove this using right/left hand limits?

Limits are DEFINED by the epsilon/delta definition. That is to say, whenever we take limits we are using the definition implicitly, and in the teaching of calculus, before limits are defined this way, the limit is not defined at all. We can only say things like "f(x) gets really really close to L as x gets closer and closer to a" which isn't very formal, so this is essentially a formalization of the above statement.

Hopefully everything I've said it correct and helps you understand a bit!

3. Originally Posted by lilaziz1
Hey everyone. I have a question on the definition of limits basically. First, the definition states:
----------------------------------------
Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then we say taht the limit of f(x) as x approaches a is L, and we write
$\lim_{x\rightarrow a} f(x) = L$

if for every number $\epsilon > 0$ there is a corresponding number $\delta > 0$ such that

if $0 < |x - a| < \delta$ then $|f(x) - L| < \epsilon$
-------------------------
So the definition is basically saying that if the difference between x -a (whether x is approaching from the left/right since its abs value bar) is approaching 0, then the difference between f(x) - L is approaching 0. Is that correct?

Now an example:

Prove that $\lim_{x\rightarrow3} (4x - 5) = 7$

Solution:

So we say: if
$0 < |x - 3| < \delta$ then $|(4x - 5) - 7| < \epsilon$

$|4x - 12| < \epsilon$

$4|x - 3| < \epsilon$

$0 < |x - 3| < \frac{\epsilon}{4}$

$0 < |x - 3| < \delta$

And that's the end of it. I don't see how that proves anything besides that delta and epsilon are proportional. How does this prove anything?
Saying that "delta and epsilon are proportional" says that delta exists- and that's the whole point!

Another question: What is the point of epsilon/delta? Can't you just prove this using right/left hand limits?

Sure you could use "right/left hand limits". And how are those defined? Using epsilon/delta, right?
There is no difference between defining the limit in terms of "right/left hand limits" and in terms of "epsilon/delta" except that in "right/left hand limits", the use of "epsilon/delta" is pushed back into the "left/right" definition.

Of course, in finding limits you seldom have to resort to "epsilon/delta"- that is used to give a rigorous definition of "limit".

4. So all you have to do is get the two equalities to be the same? If it's the same, then you proved it because the difference between the actual value of f(x) and L is ... I don't know how to finish this sentence because I don't know how to relate delta and episolon :\.

Also, what's P and Q?

5. Originally Posted by lilaziz1
So all you have to do is get the two equalities to be the same? If it's the same, then you proved it because the difference between the actual value of f(x) and L is ... I don't know how to finish this sentence because I don't know how to relate delta and episolon :\.

Also, what's P and Q?
??? P and Q are capital letters! Since I can find no "P" or "Q" in your original post nor either response, I don't know what else to say. What "P" and "Q" are you referring to?

6. In post #2, matt.qmar said, "so to prove P implies Q, start by assuming P and show that Q holds."

7. Oh, I see. He was stating a very general logical principle: to prove the statement "P implies Q" where P is any hypothesis and Q is any conclusion, you start by asserting whatever it is that P says, then use that to show that what Q says is true.