0
$\begingroup$

I am having trouble proving that the horizontal asymptotes of the function $f(x)=C/1+Ae^{-bx}$ are $y=0$ and $y=C$. The approach I am going for is to use limits such that x approaches negative/positive infinity but I am not sure how to use it to show that the horizontal asymptotes are the ones mentioned before. Assuming that the variables C, A and b are positive constants.

$\endgroup$
  • $\begingroup$ for what stand the variables,$$C,A,b$$? $\endgroup$ – Dr. Sonnhard Graubner Mar 27 '17 at 12:43
  • $\begingroup$ Assume they are positive constants $\endgroup$ – Son Jerm Mar 27 '17 at 12:44
0
$\begingroup$

then we gat $$\lim_{x\to \infty}\frac{C}{1+Ae^{-bx}}=C$$ and $$\lim_{x\to -\infty}\frac{C}{1+Ae^{-bx}}=0$$ by the limit rules

$\endgroup$
  • $\begingroup$ Could you explain to me what the limit rules you are referring to are? $\endgroup$ – Son Jerm Mar 27 '17 at 12:55
  • $\begingroup$ we have $$\lim_{x\to +\infty}e^{-bx}=0$$ if $$b>0$$ and $$\lim_{x\ to-\infty}e^{-bx}=\infty$$ if $$b>0$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 27 '17 at 12:57
  • $\begingroup$ Okay now I understand the concept of it, but could you show me the whole working out of it? $\endgroup$ – Son Jerm Mar 27 '17 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.