Answer
(b)
Work Step by Step
Let $y=f(x)=\frac{C}{1+e^{-ax}}$
$\lim\limits_{x \to \infty}f(x) = C\lim\limits_{x \to \infty} \frac{1}{1+e^{-ax}} = C\frac{1}{1+(→0)} = C$
And, $\lim\limits_{x \to -\infty}f(x) = C\lim\limits_{x \to -\infty} \frac{1}{1+e^{-ax}} = C\frac{1}{1+(→\infty)} = 0$
Both of these features are displayed by -and only displayed by- the graph (b)