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anonymous

  • one year ago

While I can see that this equation must approach 1 as x approaches infinity. How would you go about notating and using limits to say so? y[x] = E^(rx) E^(bk) / (E^(rx) E^(bk)-1)

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  1. anonymous
    • one year ago
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    \[y[x] = \frac{E^{rx} E^{bk} }{ (E^{rx} E^{bk}-1)}\]

  2. anonymous
    • one year ago
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    what are the values of \(E\) ?

  3. anonymous
    • one year ago
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    sorry.. E is the natural euler exponent ,

  4. anonymous
    • one year ago
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    is that supposed to be lower case?

  5. anonymous
    • one year ago
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    ok, and how about \(r\)? Is it a positive number?

  6. anonymous
    • one year ago
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    r is positive, and I think we can say b is too

  7. anonymous
    • one year ago
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    lets say they all are.. k too

  8. anonymous
    • one year ago
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    then\[\lim_{x \to \infty} \frac{e^{rx} e^{bk} }{ e^{rx} e^{bk}-1}=\lim_{x \to \infty} \frac{e^{rx} e^{bk} }{ e^{rx} e^{bk}}=1\]you can neglect \(1\) in the denom, because \(e^{rx} e^{bk}>>1\)

  9. anonymous
    • one year ago
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    Am I right?

  10. anonymous
    • one year ago
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    looks good to me :) so the constant 1, can be eliminated on grounds that the first terms are going to be insanely greater ... is there a need to cancel anything from that position?

  11. anonymous
    • one year ago
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    emm, no, It's fine from there

  12. anonymous
    • one year ago
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    so is this a general rule that constants are considered 0?

  13. anonymous
    • one year ago
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    yeah, you can use that fact

  14. anonymous
    • one year ago
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    sweet thank you.

  15. anonymous
    • one year ago
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    np

  16. Michele_Laino
    • one year ago
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    here, we can divide both numerator and denominator, by e^(rx), so we get: \[\Large y\left( x \right) = \frac{{{e^{rx}}{e^{bk}}}}{{{e^{rx}}{e^{bk}} - 1}} = \frac{{{e^{bk}}}}{{{e^{bk}} - {e^{ - rx}}}}\] Now, if r>0, then we have: \[\Large \mathop {\lim }\limits_{x \to + \infty } {e^{ - rx}} = 0\] then the requested limit value is: \[\Large \mathop {\lim }\limits_{x \to + \infty } y\left( x \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{{e^{bk}}}}{{{e^{bk}} - {e^{ - rx}}}} = \frac{{{e^{bk}}}}{{{e^{bk}}}} = 1\]

  17. Michele_Laino
    • one year ago
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    provided that: \[\Large r > 0\]

  18. anonymous
    • one year ago
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    thank you michele .. that's exactly what I was looking for.

  19. anonymous
    • one year ago
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    I knew there was some kind of cancellation I could use, but I couldn't see it.

  20. Michele_Laino
    • one year ago
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    :)

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