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Proving a limit

  1. Sep 15, 2007 #1
    [SOLVED] Proving a limit

    [tex]\lim_{x \rightarrow 5} \frac{x^3-6x^2+15x-47}{\{x^2-8x+14}=-3}[/tex]

    So I know how to prove a limit using epsilon and delta. The only thing is my book doesn't show how to do it with a fraction. So i get stuck at a certain point.

    I get up to [tex]\frac{x-5\times (x^2+2x+1)}{\{x^2-8x+14}\leq\epsilon}[/tex]

    But then what do i do with the numerator [tex]{(x^2+2x+1)}[/tex]

    and the denominator [tex]{(x^2-8x+14)}[/tex]

    Also I' am not sure if i restrict x to [tex]\frac{9}{\{2}\leq x \leq\frac{11}{\{2}[/tex]

    Basically I need someone to point me in the right direction about what to do with the numerator and denominator. Thanks for any help given.

    P.S : Ignore the symbol in front of the x^2 and the 2 in the denominator( I didnt know how to get rid of it). I also didnt put the absolute value where they go because i didn't know how but hopefully you guys understand. If not then thanks anyway
  2. jcsd
  3. Sep 15, 2007 #2
    If I posted something wrong or if it's not clear just tell me and i will try and change it so it makes sense.
  4. Sep 15, 2007 #3


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    Been a while since I've done epsilon delta proofs... but I'll try to help. you want |f(x) - L|<epsilon...

    what expression do you get for |f(x) - L| ?
  5. Sep 15, 2007 #4
    thanks for the reply. Well i did the f(x)-L. What i got was the fraction i showed above. This is after i combined them and factored out the x-5 to get the (x- a) is less than or equal to delta. Basically the thing below. What I don't know is what to do after this.

    [tex]\frac{x-5\times (x^2+2x+1)}{\{x^2-8x+14}\leq\epsilon}[/tex]
  6. Sep 15, 2007 #5


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    Oops... never mind... let me have a look. sorry.
    Last edited: Sep 15, 2007
  7. Sep 15, 2007 #6


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    Yes, looks like you're almost there... I think the trick is to find the maximum value of [tex]|\frac{x^2+2x+1}{x^2-8x+14}|[/tex]... or if we can say it is bounded by some maximum value then you can finish...

    Not sure how exactly...
  8. Sep 15, 2007 #7
    The trick is to pick a temporary delta neighborhood (remember your temporary delta) and bound your function. Bound the numerator and denominator separately, remembering the archemidian principle for the denominator. Youll come up delta being less than epsilon times a coefficient as usual. But you have to cover that the temporary delta inequality from the start is always secured regardless of epsilon. So your choice of delta winds up being the minimum of a number and a function of epsilon (hope this makes some sense)
  9. Sep 15, 2007 #8
    I think if restrict the x to 9/2 -11/2 interval because i think i need it to be below the graph and above the x- axis. So Then if i choose 9/2 as the value and what i get is 30.25 over 1.75. So what i think the answer is...

    [tex]\delta= min(\frac{1}2},{.057377\epsilon)[/tex]

    But I'm not sure if it's correct and can't quite prove it
  10. Sep 15, 2007 #9
    Sorry I never learned this. All i know is to restrict x to an interval and then choose (depending on the problem) the higher number and then plug it into x to get the epsilon. My Teacher doesn't explain things to clearly and it's not in my textbook so I'm at a loss...

    Edit: I think it's the same thing as I did up top but once again I'm not really sure. So any help would be appreciated
    Last edited: Sep 15, 2007
  11. Sep 15, 2007 #10


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    Hmmm... ok we can restrict x as much as we want...

    so suppose we restrict |x-5|<1/2

    ie 4.5<x<5.5...

    from your equation:


    so we need the maximum value of [tex]|\frac{x^2+2x+1}{x^2-8x+14}|[/tex]... from
    4.5<x<5.5, we can choose the maximum of the numerator magnitude wise divided the minimum of the denominator magnitude wise... and we'll know that the maximum of [tex]|\frac{x^2+2x+1}{x^2-8x+14}|[/tex] is less than that...
  12. Sep 15, 2007 #11
    This is what i did but i chose the minimum, meaning 4.5 because i thought that we need lower interval. So what i did then was substitute the 4.5 in the numerator and denominator. Then i got the answer was around .057377 times epsilon

    So if i understand you correctly i choose max for numerator and min for the denominator which makes more sense. But then how do i prove it?
  13. Sep 15, 2007 #12
    Alright when i solve the equation [tex]|x-5|\le\frac{\epsilon}{|\frac{x^2+2x+1}{x^2-8x+14}|}[/tex]

    The way you said i get [tex]|x-5|\le\\epsilon\times.04142}}[/tex]

    which i believe is right. Now i just don't understand how to prove it. Suppose i substitute the epsilon times .04142 as delta. How would i go about to show the original equation.

    P.S. : I hope you understand what I mean at least sort of. Basically how would i prove this is in fact the answer
    Last edited: Sep 15, 2007
  14. Sep 15, 2007 #13


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    So for maximum of the numerator I get |(5.5)^2 + 2(5.5) + 1| = 42.25

    For the minimum of the denominator I get |(4.5)^2 - 8(4.5) + 14| = 1.75

    I get delta as 0.0414epsilon...

    I think any interval that doesn't include the zeros of x^2-8x+14 would be fine...

    Anyway, to prove that the limit is -3, you have to work the other way right? Given an epsilon > 0. show that for delta = min{1/2, 0.0414epsilon} where |x-5|<delta, |f(x) + 3|<epsilon.

    So start with |f(x) + 3|... and show that for |x-5|<delta for the given delta... we get |f(x)+3|<epsilon

    so you'll have two cases... 1/2 <=0.0414epsilon, so delta =1/2 in this case.... and in the second case 1/2>0.414epsilon and delta =0.414epsilon.
  15. Sep 15, 2007 #14


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    Actually, I think both cases can be handled together...
  16. Sep 15, 2007 #15
    Alright the part i still don't understand is when i plug in the delta 1/2 into |x-5|<delta how do we get it to the |f(x)+3|<epsilon
  17. Sep 15, 2007 #16
    SO when i plug in 1/2 as delta. |x-5|< 1/2 how do i show that it's the same as |f(x)+3|<epsilon
  18. Sep 15, 2007 #17


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    we're given an epsilon... we choose delta = min{1/2, 0.0414epsilon} now we want to show that:

    [tex]|\frac{x^3-6x^2+15x-47}{x^2-8x+14}+3|[/tex] is less than epsilon when |x-5|<delta, for our delta.

    this equals


    we know that for |x-5|<delta... where delta = min{1/2, 0.0414epsilon}, |x-5|<1/2 (since delta is always less than or equal to 1/2) AND |x-5|<0.0414epsilon for the same reason.

    we also know that the maximum value of [tex]|\frac{x^2+2x+1}{x^2-8x+14}|[/tex] is less than (42.25/1.75) as we showed before (we showed it true for |x-5|<1/2... now we either have the same interval or a smaller one so the maximum value is either the same or smaller)...

    so [tex]|\frac{(x-5)(x^2+2x+1)}{x^2-8x+14}|[/tex] < |x-5|*(max vale of [tex]|\frac{x^2+2x+1}{x^2-8x+14}|[/tex]) < |x-5|(42.25/1.75)

    we know that |x-5|<0.0414epsilon, so |x-5|(42.25/1.75) < 0.0414epsilon(42.25/1.75)<epsilon
    Last edited: Sep 15, 2007
  19. Sep 15, 2007 #18


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    might be better to use (1.75/42.25 )epsilon instead of 0.0414epsilon... just cleaner to use exact numbers...
  20. Sep 15, 2007 #19
    i still don't get what i have to do for the proof. I know we have to work backwards but when we substitute the x for the max and min we get rid of the equation so i don't see how we get it if we go backwards.

    From |x-5|<0.0414epsilon i go to |x-5|(42.25/1.75)< epsilon

    but then how do i get it to look like the original equation [tex]|\frac{x^3-6x^2+15x-47}{x^2-8x+14}+3|[/tex] is less than epsilon
    Last edited: Sep 15, 2007
  21. Sep 15, 2007 #20


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    I proved it in post #17... can you describe which part(s) are confusing... how do you normally do epsilon delta proofs?
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