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

Nov 9, 2016

There are no vertical asymptotes, and two horizontal asymptotes at $y = 0$ and $y = 1$.

#### Explanation:

Vertical asymptotes of a rational function such as this one could occur where the function's denominator equals $0$.

Setting the denominator equal to $0$, vertical asymptotes could occur when $1 + {e}^{x} = 0$, or when ${e}^{x} = - 1$. However, we see this will never happen since the range of the function ${e}^{x}$, or any exponential function in the form ${a}^{x}$, is $y > 0$.

So, we've determined there are no vertical asymptotes. However, this doesn't account for horizontal asymptotes. We can determine these by finding the limit of the function as it approaches infinity.

Another important thing to remember is that, horizontally, we have two infinities, that is, positive and negative infinities. So, we have to take the limit at both of these values.

At positive infinity:

${\lim}_{x \rightarrow \infty} {e}^{x} / \left(1 + {e}^{x}\right) = {\lim}_{x \rightarrow \infty} \frac{1}{\frac{1}{e} ^ x + 1} = \frac{1}{0 + 1} = 1$

So, there is a horizontal asymptote at $y = 1$.

At negative infinity, we see that ${e}^{x}$ will approach $0$:

${\lim}_{x \rightarrow - \infty} {e}^{x} / \left(1 + {e}^{x}\right) = \frac{0}{1 + 0} = 0$

Our other horizontal asymptote is at $y = 0$.

Check a graph of the function!

graph{e^x/(1+e^x) [-13.7, 11.61, -1, 2]}