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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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)

Mathematics
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\[y[x] = \frac{E^{rx} E^{bk} }{ (E^{rx} E^{bk}-1)}\]
what are the values of \(E\) ?
sorry.. E is the natural euler exponent ,

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is that supposed to be lower case?
ok, and how about \(r\)? Is it a positive number?
r is positive, and I think we can say b is too
lets say they all are.. k too
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 \(e^{rx} e^{bk}>>1\)
Am I right?
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?
emm, no, It's fine from there
so is this a general rule that constants are considered 0?
yeah, you can use that fact
sweet thank you.
np
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: \[\Large \mathop {\lim }\limits_{x \to + \infty } y\left( x \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{{e^{bk}}}}{{{e^{bk}} - {e^{ - rx}}}} = \frac{{{e^{bk}}}}{{{e^{bk}}}} = 1\]
provided that: \[\Large r > 0\]
thank you michele .. that's exactly what I was looking for.
I knew there was some kind of cancellation I could use, but I couldn't see it.
:)

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