# `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))`

as `x->oo` then the `x^3/(e^(x/2))=oo/oo` form

so upon applying the L 'Hopital rule we get the solution as follows,

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))`

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

= `lim_(x->oo) (3x^2)/((e^(x/2))(1/2))`

again `(3x^2)/((e^(x/2))(1/2))` is of the form `oo/oo` so , applying the L'Hopital Rule we get

= `lim_(x->oo) (3x^2)/((e^(x/2))(1/2))`

= `lim_(x->oo) ((3x^2)')/(((e^(x/2))(1/2))')`

=`lim_(x->oo) ((6x))/(((e^(x/2))(1/2)(1/2)))`

=`lim_(x->oo) ((6x))/(((e^(x/2))(1/4)))`

again `((6x))/(((e^(x/2))(1/4))) ` is of the form `oo/oo` so , applying the L'Hopital Rule we get

=`lim_(x->oo) ((6x))/(((e^(x/2))(1/4)))`
=`lim_(x->oo) ((6x)')/(((e^(x/2))(1/4))')`
=`lim_(x->oo) ((6))/(((e^(x/2))(1/4)(1/2)))`
=`lim_(x->oo) ((6))/(((e^(x/2))(1/8)))`
=`lim_(x->oo) ((6*8))/(((e^(x/2))))`
=`lim_(x->oo) ((48))/(((e^(x/2))))`
upon plugging the value of `x= oo`
we get
=`lim_(x->oo) ((48))/(((e^((oo)/2))))`
=`lim_(x->oo) ((48))/(oo)`
=`0`