## 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)

1. anonymous

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

2. anonymous

what are the values of $$E$$ ?

3. anonymous

sorry.. E is the natural euler exponent ,

4. anonymous

is that supposed to be lower case?

5. anonymous

ok, and how about $$r$$? Is it a positive number?

6. anonymous

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

7. anonymous

lets say they all are.. k too

8. anonymous

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

Am I right?

10. anonymous

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

emm, no, It's fine from there

12. anonymous

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

13. anonymous

yeah, you can use that fact

14. anonymous

sweet thank you.

15. anonymous

np

16. Michele_Laino

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

provided that: $\Large r > 0$

18. anonymous

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

19. anonymous

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

20. Michele_Laino

:)