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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