Question

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)

Answer 1

\[y[x] = \frac{E^{rx} E^{bk} }{ (E^{rx} E^{bk}-1)}\]

Answer 2

what are the values of \(E\) ?

Answer 3

sorry.. E is the natural euler exponent ,

Answer 4

is that supposed to be lower case?

Answer 5

ok, and how about \(r\)? Is it a positive number?

Answer 6

r is positive, and I think we can say b is too

Answer 7

lets say they all are.. k too

Answer 8

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 \(e^{rx} e^{bk}>>1\)

Answer 9

Am I right?

Answer 10

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?

Answer 11

emm, no, It's fine from there

Answer 12

so is this a general rule that constants are considered 0?

Answer 13

yeah, you can use that fact

Answer 14

sweet thank you.

Answer 15

np

Answer 16

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:
\[\Large \mathop {\lim }\limits_{x \to + \infty } y\left( x \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{{e^{bk}}}}{{{e^{bk}} - {e^{ - rx}}}} = \frac{{{e^{bk}}}}{{{e^{bk}}}} = 1\]

Answer 17

provided that:
\[\Large r > 0\]

Answer 18

thank you michele .. that's exactly what I was looking for.

Answer 19

I knew there was some kind of cancellation I could use, but I couldn't see it.

Answer 20

:)