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Constant k and limits

  1. Sep 20, 2006 #1
    Im stuck on one problem in particular that I cannot figure out how to start but once I get past the first step I can figure it out. The problem is...
    lim as x-->infinity (3^kx + 6) / (3^2x + 4) and you must find a value of constant k such that the limit exists. I was thinking that k might be 2 so that the entire terms could cancel out.
     
  2. jcsd
  3. Sep 20, 2006 #2

    radou

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    Try dividing the fraction with 3^(kx).
     
  4. Sep 20, 2006 #3
    now when you divide 3^2x by 3^kx, that brings you to 3^2x-kx correct? Now do you factor out x to make it 3^x(2-k) in which case the entire equation would become 1 / 3^x(2-k) after substituting infinity into the other terms but I dont understand what to do with this last term.
     
  5. Sep 20, 2006 #4

    radou

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    Okay, start with [tex]lim_{x \rightarrow \infty} \frac{3^{kx}+6}{3^{2x}+4}=lim_{x \rightarrow \infty} (\frac{3^{kx}}{3^{2x}+4}+\frac{6}{3^{2x}+4})=lim_{x \rightarrow \infty}\frac{3^{kx}}{3^{2x}+4}+lim_{x \rightarrow \infty} \frac{6}{3^{2x}+4}=lim_{x \rightarrow \infty}\frac{3^{kx}}{3^{2x}+4}[/tex]. Now divide with 3^(kx) and factor out x, as mentioned above.
     
    Last edited: Sep 20, 2006
  6. Sep 20, 2006 #5
    after following that I came out to have 1 / 3^x(2-k) ... is that what you mean by factoring out x? However I still dont understand what the value of k would be if i still have x in the equation. Maybe im just not seeing something..
     
  7. Sep 20, 2006 #6

    radou

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    You now have [tex]lim_{x \rightarrow \infty} \frac{1}{3^{x(2-k)}+\frac{4}{3^{kx}}}[/tex]. All you have to do now is to see if there will be a limit for different values of k. The interesting cases are k = 2, k > 2 and k < 2.
     
  8. Sep 20, 2006 #7

    dextercioby

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    I advise you to use the "\" before "lim" in order to look better.

    [tex] \lim_{x\rightarrow \infty} [/tex]

    C?

    Daniel.
     
  9. Sep 20, 2006 #8

    radou

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    Thanks... I just don't understand how to get the [tex]x \rightarrow \infty[/tex] below the limit. The source code is the same, I don't get it. Anyway, off topic. :)
     
    Last edited: Sep 20, 2006
  10. Sep 20, 2006 #9
    so now if k=2 the limit is 1, if k > 2 then the limit is infinity, if k < 2 then the limit is 0, did I solve that right or what am I missing now?
     
    Last edited: Sep 20, 2006
  11. Sep 20, 2006 #10
    does it mean that the limit exists at k < or = to 2?
     
  12. Sep 20, 2006 #11
    Yeah I caught my mistake for k < 2 right after I posted it but how is the limit not 1 for k = 2, when 2 is put into 3^x(2-k) that would make it 3^0 ultimately which equals 1, and 4/3^kx becomes 0
     
  13. Sep 20, 2006 #12

    radou

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    Yes, for k < 2 it exists, and it equals zero. Further on, it exists for k = 2, and it equals 1. Sorry, I messed something up. :)
     
  14. Sep 20, 2006 #13
    Alright I finally understand it now, thank you very much for the help
     
  15. Sep 20, 2006 #14
    ok one more question, if the problem was the exact same but this time approach negative infinity, I did out the problem and would the limit now exist at k > or = to 2?
     
  16. Sep 20, 2006 #15

    radou

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    If [tex]x \rightarrow -\infty[/tex], for k < 0, the limit is infinity, for k = 0 it equals 1/4, and for k > 0 it equals zero.
     
  17. Sep 20, 2006 #16
    may I just ask why you use 0 instead of 2?
     
  18. Sep 20, 2006 #17

    radou

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    Because for k = 0 'something changed'. :biggrin:
     
  19. Sep 20, 2006 #18
    you still use the same method of simplifying the equation as we did for the first question right?
     
  20. Sep 20, 2006 #19

    radou

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    Yes. In our case, it was all about bringing the function (whose limit is of interest to us) into a suitable form to apply the simple limit [tex]\lim_{x\rightarrow\infty}\frac{1}{x}=0[/tex].
     
  21. Sep 20, 2006 #20
    true true, however in the first one we were able to cancel out the other side of the formula of 6/3^2x + 4 because of that simple limit. However now you cannot because when negative infinity is put into its spot it becomes undef
     
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