# `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...

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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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