Given to solve,

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

as `x->oo` then the ` e^x/(x^4) =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) (e^x)/(x^4)`

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

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

again `((e^x))/((4x^3))` is of the form `oo/oo` so , we can apply again L'Hopital Rule .

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

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

=`lim_(x->oo) ((e^x))/((12x^2))`

again `((e^x))/((12x^2))` is of the form `oo/oo ` so , we can apply again L'Hopital Rule .=`lim_(x->oo) ((e^x)')/((12x^2)')`

=`lim_(x->oo) ((e^x))/(((12*2)x))`

= `lim_(x->oo) ((e^x))/(((24)x))`

again `((e^x))/(((24)x))` is of the form `oo/oo` so , we can apply again L'Hopital Rule .

= `lim_(x->oo) ((e^x)')/(((24)x)')`

=`lim_(x->oo) ((e^x))/(24)`

on plugging the value`x= oo` , we get

=`((e^(oo)))/(24)`

=`oo`

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