Given to solve,

`lim_(x->oo) ((5x+3)/(x^3-6x+2))`

as `x->oo` then the `((5x+3)/(x^3-6x+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)((5x+3)/(x^3-6x+2))`

= `lim_(x->oo) ((5x+3)')/((x^3-6x+2)')`

= `lim_(x->oo) (5)/(3x^2-6)`

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

=` (5)/(3(oo)^2-6)`

= `5/oo`

= `0`

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