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

Question

Serenity Jones · Accepted Answer

There is no vertical asymptote. (assuming we are restricted to the Real number plane)
Horizontal asymptotes at #y=1# and #y=0# Vertical Asymptote Since #e^x > 0# for all Real values of #x# the denominator of #(e^x)/(1+e^x)# will never be #=0# and the expression is defined for all values of #x# Horizontal Asymptote #(e^x)/(1+e^x) = ((e^x)/(e^x))/(1/(e^x)+(e^x)/(e^x)) = 1/(1/(e^x)+1)# and since #color(white)("XXX")1/(e^x)rarr 0# as # xrarr+oo# and #color(white)("XXX")(e^x)rarr0# as # xrarr-oo# #color(white)("XXX")#which implies #1/e^x rarr oo# as #xrarr -oo# therefore #color(white)("XXX")lim_(xrarr+oo) y = 1/(0+1) = 1# and #color(white)("XXX")lim_(xrarr-oo) y = 1/(oo+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]}

Serenity Johnson · Answer

undefined To find the asymptotes of the function $ \frac{e^x}{1+e^x} $, we need to examine its behavior as $ x $ approaches positive or negative infinity.

1. As $ x $ approaches positive infinity:
$$ \lim_{x \to \infty} \frac{e^x}{1+e^x} = \lim_{x \to \infty} \frac{e^x}{e^x(1+e^{-x})} = \lim_{x \to \infty} \frac{1}{1+e^{-x}} = 1 $$
So, there is a horizontal asymptote at $ y = 1 $.

2. As $ x $ approaches negative infinity:
$$ \lim_{x \to -\infty} \frac{e^x}{1+e^x} = \lim_{x \to -\infty} \frac{1}{e^{-x}+1} = 0 $$
So, there is a horizontal asymptote at $ y = 0 $.

Therefore, the horizontal asymptotes of $ \frac{e^x}{1+e^x} $ are $ y = 1 $ and $ y = 0 $.