# How to find the horizontal asymptotic of the logistic function $C/1+Ae^-{bx}$ using limits

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.

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

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

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