lim_(x->oo) (5x^2+3x-1)/(4x^2+5) Evaluate the limit, using L’Hôpital’s Rule if necessary.

Given to solve,
lim_(x->oo) (5x^2 +3x-1)/(4x^2 +5)
as x->oo then the (5x^2 +3x-1)/(4x^2 +5)=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^2 +3x-1)/(4x^2 +5)
= lim_(x->oo) ((5x^2 +3x-1)')/((4x^2 +5)')
= lim_(x->oo) (10x+3)/(8x)
the above limit is of the form oo/oo
so again applying the L'Hopital rule we get
 lim_(x->oo) (10x+3)/(8x)
= lim_(x->oo) ((10x+3)')/((8x)')
=lim_(x->oo) (10)/((8))
= 10/8
=5/4