`lim_(x->oo)x^3/e^(x^2)` Evaluate the limit, using L’Hôpital’s Rule if necessary.

Given to solve ,

`lim_(x->oo) x^3/e^(x^2)`

upon `x` tends to` oo` we get `x^3/e^(x^2) = oo/oo`

so, by applying the L'Hopital Rule we get

as for the general equation it is as follows

`lim_(x->a) f(x)/g(x) is = 0/0` or `(+-oo)/(+-oo)` then by using the L'Hopital Rule we get  the solution with the  below form.

`lim_(x->a) (f'(x))/(g'(x))`

so , now evaluating

`lim_(x->oo) x^3/e^(x^2)`

upon using the L'Hopital Rule

=`lim_(x->oo) ((x^3)')/((e^(x^2))')`

=`lim_(x->oo) (3x^2)/(e^(x^2)(2x))`

=>`lim_(x->oo) (3x)/(e^(x^2)(2))`

now...

Given to solve ,

`lim_(x->oo) x^3/e^(x^2)`

upon `x` tends to` oo` we get `x^3/e^(x^2) = oo/oo`

so, by applying the L'Hopital Rule we get

as for the general equation it is as follows

`lim_(x->a) f(x)/g(x) is = 0/0` or `(+-oo)/(+-oo)` then by using the L'Hopital Rule we get  the solution with the  below form.

`lim_(x->a) (f'(x))/(g'(x))`

so , now evaluating

`lim_(x->oo) x^3/e^(x^2)`

upon using the L'Hopital Rule

=`lim_(x->oo) ((x^3)')/((e^(x^2))')`

=`lim_(x->oo) (3x^2)/(e^(x^2)(2x))`

=>`lim_(x->oo) (3x)/(e^(x^2)(2))`

now on` x-> oo` we get `(3x)/(e^(x^2)(2)) =oo/oo`

so,again by applying the L'Hopital Rule we get

`lim_(x->oo) (3x)/(e^(x^2)(2))`

=`lim_(x->oo) ((3x)')/((e^(x^2)(2))')`

=`lim_(x->oo) (3)/(e^(x^2)(2)(2x))`

=`lim_(x->oo) (3)/(e^(x^2) (4x))`

now as `x-> oo`

`3/(e^(x^2) (4x)) =3/(e^(oo^2) (4(oo))) =0`

so

`lim_(x->oo) (3)/((e^(x^2) (4x)))` `=0`

now we can state that

`lim_(x->oo) x^3/e^(x^2)` `=0`