How do you find the asymptotes for # f(x)=xe^-x#?

1 Answer
Jun 20, 2018

Answer:

#C_f# has an horizontical asymptote #y=0# at #+oo#

Explanation:

#f(x)=xe^(-x)=x/e^x#

  • #f# has domain the set of real numbers #RR# because #e^x>0# , #AAx##in##RR#

This means we're not looking for vertical asymptotes.

For oblique linear asymptotes #(y=λx+b)# we have:

#lim_(xto-oo)f(x)/x=lim_(xto-oo)(xe^(-x))/x=lim_(xto-oo)e^(-x)=#

#+oo#

No oblique/horizontical asymptote at #-oo#

#lim_(xto+oo)f(x)/x=lim_(xto+oo)e^(-x)=^((e^(-oo)))0=λ#

#lim_(xto+oo)(f(x)-λx)=lim_(xto+oo)f(x)=lim_(xto+oo)xe^(-x)=#

#lim_(xto+oo)x/e^x=_(DLH)^(((+oo)/(+oo)))lim_(xto+oo)1/e^x=^((1/(+oo)))0=b#

As a result #C_f# has an horizontical asymptote #y=0# at #+oo#