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

1 Answer
Feb 3, 2016

Answer:

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 #(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]}