Question

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

Answer 1

`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`