Proove A Limit Using Delta and Epsilon

In summary, the conversation revolved around finding the limit as x approaches 4 for the function f(x)=x^2-8x-16. The conversation touched on the definition of precise limits and using factoring to simplify the expression. The final answer was found to be the square root of e.
  • #1
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Homework Statement



Proove the limit as x approaches 4 for f(x)=x^2-8x= -16

Homework Equations



Definition of Precise Limits

The Attempt at a Solution



I know that I want x^2-8x+16 (after moving the -16 over per the limit definition) to look like |x-4|

Factoring gets me (x-4)(x-4)<e

Because I have two factors, I want to bound one of them so that -1<|x-4|<1.

Adding the four to both sides, I get 3<x<5, but when I add four back in (since both factors are the same), I'm back to -1<|x-4|<1. This leads me to believe that C=1, but I don't think I'm doing this part right. I know the answer isn't 'e'.

I'm lost at this point.
 
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  • #2
Ok, you have x2-8x+16=(x-4)2.

Now the objective is: Given e>0, what must |x-4| be smaller than to make |x-4|2<e? You can usually do this backwards, that is start with |x-4|2<e and see if you can find an inequality for |x-4| from this
 
  • #3
That's what I thought I was doing, but I know I'm not getting the correct answer. I think I'm looking for an E/C, but when C=1, it's the same as E.
 
  • #4
I'm confused by what C is supposed to be.

If you have |x-4|2<e, what operation should you do to "solve" for |x-4|?
 
  • #5
From what I understand, if you have two factors, but only one matches the inequality for the limit (in this case -1<|x-4<1 )...I need to find a positive constant C such that the absolute value of the other factor is <C.

This is a bit confusing since the original equation factors out to (x-4)(x-4). They are the same.

To do this, I bound the factor I want which can be -1<|x-4|<1. Adding the x to both sides makes it 3<x<5. To get C, I take the other factor and apply it to this new inequality. The other factor is also x-4 so I subtract 4 from the x in 3<x<5 which takes me right back to -1<x-4<1. This means that C equals 1 and the two restrictions on the factor I need is |x-4|<1 and |x-4|<e/C. That is e/1 which is only e.

This is wrong.

Sorry for the confusion, I'm still trying to get a clear understanding of the subject myself.
 
  • #6
Guh...I figured it out after I walked away from the problem for awhile. (x-4)(x-4) is (x-4)^2. That means I only had to take the square root of both sides and the answer is sqrt of e.

No wonder you were asking me about C...it wasn't even needed for this.

Thank you for showing me the problem in a different way. :) It helped me figure it out.
 
  • #7
"Moving the 16 over", in f(x)= x^2- 8x= 16 gives x^2- 8x- 16, not x^2- 8x+ 16.
 
  • #8
That is my fault; the original problem has -16, not 16. I fixed my original post. Thank you for pointing that out.
 

FAQ: Proove A Limit Using Delta and Epsilon

What is the definition of a limit using delta and epsilon?

The definition of a limit using delta and epsilon is a mathematical method used to formally prove that a function approaches a certain value as its input approaches a given value. It involves finding a suitable value for delta (denoted by δ) and epsilon (denoted by ɛ) that satisfies a specific set of conditions.

How do you prove a limit using delta and epsilon?

To prove a limit using delta and epsilon, you must first set up the definition of a limit by stating that for any epsilon value greater than 0, there exists a corresponding delta value that satisfies a specific condition. Then, you must manipulate the equation to isolate the delta value and find a suitable expression for it. Finally, you must choose a value for delta that satisfies the original condition and show that it also satisfies the modified equation.

What is the purpose of using delta and epsilon in proving a limit?

The use of delta and epsilon in proving a limit helps to provide a rigorous and formal way of showing that a function approaches a certain value as its input approaches a given value. It helps to remove any ambiguity or doubt in the proof and ensures that the limit is true for all possible values of the input.

Can you explain the role of delta and epsilon in the definition of a limit?

In the definition of a limit, delta and epsilon play specific roles. Delta represents the distance between the input and the limit point, while epsilon represents the acceptable margin of error. The condition for the limit to hold is that the distance between the input and the limit point must be less than the margin of error. In other words, the input must be within a certain distance of the limit point for the limit to be true.

What are some common mistakes when proving a limit using delta and epsilon?

Some common mistakes when proving a limit using delta and epsilon include choosing an incorrect value for delta, not showing that the chosen delta satisfies the original condition, and using incorrect algebraic manipulations. It is also important to keep in mind that a limit can only be proven using delta and epsilon if it exists, so one should always check for the existence of the limit before attempting to prove it.

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