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Limit of a rational function for different values of K (check)

  1. Jun 24, 2011 #1
    Find the limit of (4kx + 9)/(45x + 3) as x->inf, when K=5, K<5, K>5.

    Answer: I divided each term by the highest power of 4, and was able to come up with 0 (K<5), 1 (K=5), inf (K>5). I'm pretty sure these are correct.

    Was this the right method? Am I good to assume that dividing by the highest power of x is similar in some ways to dividing by the highest power of some common base, which is what I did here?

    Let me know if I can explain myself better.
     
  2. jcsd
  3. Jun 24, 2011 #2

    gb7nash

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    You are indeed correct. If you want to be more rigorous for K = 5, K > 5, and 0 <= k < 5 use the following trick:

    [tex]\frac{4^{kx} + 9}{4^{5x} + 3} = \frac{4^{kx} + 9}{4^{5x} + 3} \ * \frac{\frac{1}{4^{kx}}}{\frac{1}{4^{kx}}} = \frac{1 + 9*4^{-kx}}{4^{(5-k)x} + 3*4^{-kx}} [/tex]

    For k < 0, you can use what you originally typed (do you know why?)
     
    Last edited: Jun 24, 2011
  4. Jun 24, 2011 #3
    Thanks.

    And for K < 0, I can use what I originally typed because the terms with x in their exponent will tend to zero as x -> inf.

    And how do you type those operations out? They look so pretty.
     
  5. Jun 24, 2011 #4

    micromass

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    Yes, for K<0, we have that [itex]4^{Kx}\rightarrow 0[/itex]. And thus

    [tex]\frac{4^{Kx}+9}{4^{5x}+3}\rightarrow 0[/tex]

    as the denominator will tend to infinity.

    Click on the equations to see the code! :smile:
     
  6. Jun 24, 2011 #5

    gb7nash

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  7. Jun 24, 2011 #6
    micromass: I was pretty general (and wrong) with my reply. Thanks for the clarification.
     
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