Walkthrough of Epsilon Delta Limits Question

In summary, if 0<|x-c|<delta, then |f(x) - L|< Epsilon. If |x-3|<1, then 2<x<4 and |x+3|<=|x|+|3|= x+3<7.
  • #1
Live4eva_2
28
0
Could someone please give me a walkthrough of the following question(and answer)??
I really can't understand it...

lim x^2 = 9
x->3

if 0<|x-c|<delta then |f(x) - L|< epsilon
Ok,I understand that...I need x-3<delta and x^2 -9<epsilon

so... x^2 - 9 = (x+3)(x-3)
factorise...

|x^2 - 9| = |x+3||x-3|


Here's the problem.The book states:

We need to get an estimate for |x+3| close to 3.For convenience we'll
take within one unit of 3.
An estimate of what??And how do I know when I must estimate??I'm guessing that it doesn't matter whether (x+3) or (x-3) is chosen due to the absolute value brackets..?
anyway:
if |x-3|<1 then 2<x<4 (Choosing a delta...now x lies between 2 and 4.)
|x+3|<=|x|+|3|= x+3<7 (this would be: c-delta<x<c+delta,correct?) **How do I know when to implement that ??**Why are they now using |x+3| instead of (x-3)?!?

if |x-3|<1 then |x^2 -9|< 7|x-3|
I understand that this solution basically states delta = 1/7 epsilon??And I know that the 7 is acquired through the c-delta<x<c+delta formula/method but I'm still too lost!

please clarify guys!
 
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  • #2
Live4eva_2 said:
Could someone please give me a walkthrough of the following question(and answer)??
I really can't understand it...

lim x^2 = 9
x->3

if 0<|x-c|<delta then |f(x) - L|< epsilon


so... x^2 - 9 = (x+3)(x-3)


|x^2 - 9| = |x+3||x-3|
Good so far. You want [itex]|x+3||x-3|< \epsilon[/itex] and you have to make sure that is true when [itex]|x-3|< \delta[/itex]. Notice that one has |x+3| and the other doesn't!


Here's the problem.The book states:
"We need to get an estimate for |x+3| close to 3.For convenience we'll
take within one unit of 3. "

An estimate of what??And how do I know when I must estimate??I'm guessing that it doesn't matter whether (x+3) or (x-3) is chosen due to the absolute value brackets..?
An estimate of how large |x+3| is close to 3. And it certainly does matter "whether (x+3) or (x-3) is chosen". You are taking the limit as x-> 3, not -3! You want get [itex]|x-3|< \delta[/itex], not [itex]|x+3|< \delta[/itex].


anyway:
if |x-3|<1 then 2<x<4 (Choosing a delta...now x lies between 2 and 4.)
|x+3|<=|x|+|3|= x+3<7 (this would be: c-delta<x<c+delta,correct?) **How do I know when to implement that ??**Why are they now using |x+3| instead of (x-3)?!?

if |x-3|<1 then |x^2 -9|< 7|x-3|
I understand that this solution basically states delta = 1/7 epsilon??And I know that the 7 is acquired through the c-delta<x<c+delta formula/method but I'm still too lost!

please clarify guys!

You want [itex]|x-3||x+3|< \epsilon[/itex] whenever [itex]|x-3|< \delta[/itex] and the question is, how should you choose [itex]\delta[/itex].
You could try writing [itex]|x- 3|< \epsilon/|x+3|[/itex] but that depends on x, it is not a constant. That's where the "estimate for |x+3|" comes in. You need to replace |x+3| by a constant, K, and be sure that K> |x+3| so that [itex]\epsilon/K< \epsilon/|x+3|[/itex].

Since we are going to be "close to 3" anyway, suppose |x-3|< 1- that is, -1< x-3< 1. Adding 3 to both sides, 2< x< 4 and, adding another 3, 5< x+3< 7 so if |x-1|< 1, |x+3|< 7. Then 1/7< 1/|x+3| (notice how the inequality swaps direction when we put the numbers in the denominator: 2< 3 so 1/3< 1/2). Since 1/7< 1/|x+3|, [itex]\epsilon/7< \epsilon/|x+3|[/itex]. We can choose [itex]\delta[/itex] equal to the smaller of 1 and [itex]\epsilon/7[/itex] so they are both true.

If [itex]|x-3|< \delta< \epsilon/7< \epsilon/|x+3|[/itex], then, multiplying both sides by the positive number |x+3|, [itex]|x-3||x+ 3|= |x^2- 9|< \epsilon[/itex].

(Of course, the "|x-3|< 1" is just convenient. |x-3|< any fixed positive number would work as well.)
 
  • #3
OK I'm going to print this out in the morning and get back to you on this one.(in the South African morning that is)Thanks...
 
  • #4
OK,I had a look last night.I had problems with the following:

I didn't realize that the value x was approaching(in this case 3),had any other function than to be plugged in |x-c|<delta.So I learned something there.But :
so if |x-1|< 1, |x+3|< 7. Then 1/7< 1/|x+3|

I don't know where |x-1| came from...

Then 1/7< 1/|x+3| (notice how the inequality swaps direction when we put the numbers in the denominator: 2< 3 so 1/3< 1/2). Since 1/7< 1/|x+3|, . We can choose equal to the smaller of 1 and so they are both true.

I can see that Epsilon is 7 times bigger than delta(in Salas) but I don't quite understand this method of making the inequality into a fraction...

And just to check:
0<|x-c|<delta is equivalent to c-delta<x<c+delta and can be interchanged at any time??
 
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  • #5
Live4eva_2 said:
OK,I had a look last night.I had problems with the following:

I didn't realize that the value x was approaching(in this case 3),had any other function than to be plugged in |x-c|<delta.So I learned something there.But :


I don't know where |x-1| came from...
That was a typo. I meant, |x-3|, of course.

I can see that Epsilon is 7 times bigger than delta(in Salas) but I don't quite understand this method of making the inequality into a fraction...
If 0< a< b, (and neither is 0) then "taking the reciprocal" reverses the inequality: 1/a> 1/b. You can do that step by step: first divide both sides by the positive number a: 1< b/a, and then divide both sides by the positive number b: 1/b< 1/a which is the same as 1/a> 1/b.

And just to check:
0<|x-c|<delta is equivalent to c-delta<x<c+delta and can be interchanged at any time??
Yes, that follows from the definition of absolute value. If x>= c, then x-c is non-negative so |x-c|= x- c: x- c< delta => x< c+ delta. If x< c, then x- c is negative so |x- c|= -(x-c): -(x-c)< delta=> x-c> -delta=> x> x- delta.
 
  • #6
Ok I attempted a very similar problem to try and get a feel for this,although I have gotten a bit stuck.This is what I've done:

question: lim x^2 =4
x->2
my answer:

|x-2|<delta
|x^2 - 4|<epsilon
|x-2||x+2|<epsilon
Let |x+2|<1
-1<x+2<1
-3<x<-1 It's gone negative here...Doesn't seem right.
-5<x-2<-3
|x-2|<-3

couldn't i assume that if |x-2|<-3 then |x+2|<1?


so if |x-1|< 1, |x+3|< 7. Then 1/7< 1/|x+3|
I would have assumed that if |x-1|<1 then |x+3|<4...(by adding 4 to |x-1| to equate it with |x+3| ?)
 
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  • #7
Live4eva_2 said:
Ok I attempted a very similar problem to try and get a feel for this,although I have gotten a bit stuck.This is what I've done:

question: lim x^2 =4
x->2
my answer:

|x-2|<delta
|x^2 - 4|<epsilon
|x-2||x+2|<epsilon
Let |x+2|<1
You've got this backwards. Since the limit is take at x= 2, you want x close to 2, so x- 2 (not x+2) close to 0. For example, |x-2|< 1.

-1<x+2<1
-3<x<-1 It's gone negative here...Doesn't seem right.
-5<x-2<-3
|x-2|<-3

couldn't i assume that if |x-2|<-3 then |x+2|<1?
|x-2| is NOT "< -3", absolute value is never negative. Saying that -5< x-2< -3 tells you, first, that x-2< 0! In that case, |x-2|= -(x-2). Multiplying each part of the inequality by the negative number -1, 5> -(x-2)> 3 or 3< |x-2|< 5. |x-2|< 5.
Actually, you can get that directly from -5< x-2< -3. The larger of the two absolute values, 5 and 3, is 5: |x-2|< 5.

However, you should be thinking: Take x close to 2: say |x- 2|< 1. Then -1< x- 2< 1 so, adding 4 to each part, 3< x+ 2< 5. |x+2|< 5. Then |x-2||x+2|< 5|x-2|< [itex]\epsilon[/itex] so we need |x-2|< [itex]\epsilon/5[/itex]. Take [itex]\delta[/itex] to be the smaller of [itex]\epsilon/5[/itex] and 1.

I would have assumed that if |x-1|<1 then |x+3|<4...(by adding 4 to |x-1| to equate it with |x+3| ?)
Is this a different example? If |x-1|< 1, then -1< x- 1< 1 so 3< x+ 3< 5. |x+3|< 5.
 
  • #8
The function does not have to "pass through both corners". But you want to make the estimate as accurate as possible so you construct the rectangle that way. Where ever your "given" horizontal lines cross the graph, those are the corners of your rectangle.
 
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  • #9
OK ok...I think for now this is as good as I'm going to get it!

**But be warned!I will be spamming the forums with difference quotient problems on Monday.Let some of the other mods have a turn(-8**

Thanks Halls
 
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  • #10
additional question

Quote: Multiplying each part of the inequality by the negative number -1, 5> -(x-2)> 3 or 3< |x-2|< 5. |x-2|< 5.

Question: I am having trouble understanding the concept behind a variation of this... for example, I do understand getting into the format of -1 < (x-2) <1 then you add 2 to both sides and then you add another two to get into the format of (x+2). However what happens if you are trying to get into another format because the original equation is different and end up, for example, with
-3<(x+2)<5... meaning what happens when one side is negative and the other isn't.. is there ever a case where you end up taking the smaller of the 2 absolute values or is it always the case that it is the larger absolute value that you end up being "less than." If there is a negative value on either side, do you always multiply by -1 which could then lead to the other side being negative?

I appreciate any help with this concept.

Thank you,

jason
 

1. What is a walkthrough of epsilon delta limits?

A walkthrough of epsilon delta limits is a method used in calculus to formally prove the existence of a limit of a function. It involves using the concepts of epsilon and delta to define a range of values around a limit point, and showing that for any epsilon (small positive number), there exists a corresponding delta (small positive number) such that if the distance between the input and the limit point is less than delta, then the distance between the function output and the limit is less than epsilon.

2. Why is the epsilon delta method important?

The epsilon delta method is important because it provides a rigorous and precise way to prove the existence of limits in calculus. It allows us to confidently state that a limit exists for a particular function, rather than just making an educated guess based on the behavior of the function.

3. What are the steps involved in a walkthrough of epsilon delta limits?

The steps involved in a walkthrough of epsilon delta limits include:

  1. Identifying the limit point and the function to be evaluated.
  2. Setting up the definition of a limit using the epsilon and delta variables.
  3. Manipulating the expression to find a suitable delta value in terms of epsilon.
  4. Using algebraic manipulations and/or inequalities to simplify the expression and find a specific delta value.
  5. Proving that the limit is valid by showing that for any input within delta distance of the limit point, the output is within epsilon distance of the limit value.

4. What are some common mistakes made in a walkthrough of epsilon delta limits?

Some common mistakes made in a walkthrough of epsilon delta limits include:

  • Not clearly defining the limit point or using incorrect notation.
  • Not following the proper steps and skipping important algebraic manipulations.
  • Using epsilon and delta values that are too large or too small, resulting in an invalid proof.
  • Not considering all possible cases, such as when the input is negative or when the limit point is at infinity.

5. Can the epsilon delta method be used for all functions?

Yes, the epsilon delta method can be used for all functions that have a limit. However, it may not always be the most efficient or practical method. In some cases, other techniques such as the Squeeze Theorem or L'Hopital's Rule may be more useful in proving the existence of a limit.

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