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)))`

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