Proove A Limit Using Delta and Epsilon

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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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Answers and Replies

  • #2
Office_Shredder
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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
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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
Office_Shredder
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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
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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
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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
HallsofIvy
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"Moving the 16 over", in f(x)= x^2- 8x= 16 gives x^2- 8x- 16, not x^2- 8x+ 16.
 
  • #8
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That is my fault; the original problem has -16, not 16. I fixed my original post. Thank you for pointing that out.
 

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