Given to solve,

`lim_(x->oo) lnx/(x^2)`

as `x->oo` then the `lnx/(x^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...

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Given to solve,

`lim_(x->oo) lnx/(x^2)`

as `x->oo` then the `lnx/(x^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) lnx/(x^2)`

= `lim_(x->oo) ((lnx)')/((x^2)')`

= `lim_(x->oo) (1/x)/((2x))`

= `lim_(x->oo) (1/(2x^2))`

upon plugging in`x= oo` , we get

= `(1/(2(oo)^2))`

`= 1/oo = 0`