How do you find the asymptotes for #y = (5e^x)/ ((e^x) - 8)#?

1 Answer
Feb 24, 2017

Answer:

Vertical asymptote at #x = ln8#
Horizontal asymptotes at: #y = 0# and #y = 5#

Explanation:

The vertical asymptote is found when #D(x) = 0#:
#e^x - 8 = 0# so #e^x = 8#

Solve for #x# by logging both sides of your equation:

#ln e^x = ln 8#

Vertical asymptote at #x = ln8#

Finding horizontal asymptotes :

  1. #N(x) = 5e^x = 0; e^x = 0; ln e^x = ln 0; x = ln 0# which is undefined. By definition logarithms are positive: #x > 0# This means there is a horizontal asymptote at #y = 0#

  2. #lim x->oo (5e^x/e^x)/(e^x/e^x - 8/e^x) = 5/(1-0) = 5# so there is a horizontal asymptote at #y = 5#

Summary:
Vertical asymptote at #x = ln8#
Horizontal asymptotes at: #y = 0# and #y = 5#