# How do you find the asymptotes for (e^x)/(1+e^x)?

Feb 3, 2016

There is no vertical asymptote. (assuming we are restricted to the Real number plane)
Horizontal asymptotes at $y = 1$ and $y = 0$

#### Explanation:

Vertical Asymptote
Since ${e}^{x} > 0$ for all Real values of $x$
the denominator of $\frac{{e}^{x}}{1 + {e}^{x}}$ will never be $= 0$
and the expression is defined for all values of $x$

Horizontal Asymptote
$\frac{{e}^{x}}{1 + {e}^{x}} = \frac{\frac{{e}^{x}}{{e}^{x}}}{\frac{1}{{e}^{x}} + \frac{{e}^{x}}{{e}^{x}}} = \frac{1}{\frac{1}{{e}^{x}} + 1}$
and
since
$\textcolor{w h i t e}{\text{XXX}} \frac{1}{{e}^{x}} \rightarrow 0$ as $x \rightarrow + \infty$
and
$\textcolor{w h i t e}{\text{XXX}} \left({e}^{x}\right) \rightarrow 0$ as $x \rightarrow - \infty$
$\textcolor{w h i t e}{\text{XXX}}$which implies $\frac{1}{e} ^ x \rightarrow \infty$ as $x \rightarrow - \infty$

therefore
$\textcolor{w h i t e}{\text{XXX}} {\lim}_{x \rightarrow + \infty} y = \frac{1}{0 + 1} = 1$
and
$\textcolor{w h i t e}{\text{XXX}} {\lim}_{x \rightarrow - \infty} y = \frac{1}{\infty + 1} = 0$

For verification purposes, here's what the graph looks like:
graph{e^x/(1+e^x) [-6.24, 6.244, -3.12, 3.12]}