How do you find the asymptotes for $f (x) = x e^{- x}$ $f(x)=xe^-x$ ?

Question

9242 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-20T08:56:15

$C_{f}$ $C_f$ has an horizontical asymptote $y = 0$ $y=0$ at $+ \infty$ $+oo$

$f (x) = x e^{- x} = \frac{x}{e^{x}}$ $f(x)=xe^(-x)=x/e^x$

$f$ $f$ 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 $+oo$

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