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

1 Answer
Jun 20, 2018

C_fCf has an horizontical asymptote y=0y=0 at +oo+

Explanation:

f(x)=xe^(-x)=x/e^xf(x)=xex=xex

  • ff has domain the set of real numbers RR because e^x>0 , AAxinRR

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