Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 23

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

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