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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)
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y[x] = \frac{E^{rx} E^{bk} }{ (E^{rx} E^{bk}1)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what are the values of \(E\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry.. E is the natural euler exponent ,

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that supposed to be lower case?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, and how about \(r\)? Is it a positive number?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0r is positive, and I think we can say b is too

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lets say they all are.. k too

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then\[\lim_{x \to \infty} \frac{e^{rx} e^{bk} }{ e^{rx} e^{bk}1}=\lim_{x \to \infty} \frac{e^{rx} e^{bk} }{ e^{rx} e^{bk}}=1\]you can neglect \(1\) in the denom, because \(e^{rx} e^{bk}>>1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0looks 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0emm, no, It's fine from there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so is this a general rule that constants are considered 0?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, you can use that fact

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5here, 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: \[\Large \mathop {\lim }\limits_{x \to + \infty } y\left( x \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{{e^{bk}}}}{{{e^{bk}}  {e^{  rx}}}} = \frac{{{e^{bk}}}}{{{e^{bk}}}} = 1\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5provided that: \[\Large r > 0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you michele .. that's exactly what I was looking for.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I knew there was some kind of cancellation I could use, but I couldn't see it.
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