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

Question

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.

Answer 1

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