Find limits for `lim_{x->infty} xe^{-x}`  Use L'Hospital's Rule

2 Answers

lfryerda's profile pic

lfryerda | High School Teacher | (Level 2) Educator

Posted on

We need to rewrite the limit using algebra to get it into the form 0/0 then we can apply L'Hopital's Rule.

`lim_{x->infty} xe^{-x}`

`=lim_{x->infty}x/e^x`    now apply L'Hopital

`=lim_{x->infty} 1/e^x`


The limit is 0.

tonys538's profile pic

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

The limit `lim_(x->oo) x*e^-x` has to be determined.

l'Hospital's rule can only be applied in limits that have the indeterminate form `0/0` or `oo/oo` . There are several other indeterminate forms but they do not allow l'Hospital's rule to be applied.

`lim_(x->oo) x*e^-x`

= `lim_(x->oo) x/e^x`

When `x = oo` , `e^x = oo` . This makes `x/e^x` take the indeterminate form `oo/oo` .

Now applying l'Hospital's rule the numerator and denominator can be replaced by their derivatives.

The limit is now:

`lim_(x->oo) 1/e^x`

Substituting `x = oo` gives `1/oo = 0`

The required limit `lim_(x->oo) x*e^-x = 0`