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`lim_(x->oo)lnx/x^2` Evaluate the limit, using L’Hôpital’s Rule if necessary.

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`

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