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`lim_(x->oo) e^x/x^4` Evaluate the limit, using L’Hôpital’s Rule if necessary.


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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