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

Expert Answers

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Givne to solve ,


= `lim_(x->oo) 4 ln(x)/x^3`

= `4lim_(x->oo) ln(x)/x^3`

as `x->oo` then the `ln(x)/x^3 =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

`4lim_(x->oo) ln(x)/x^3`

=`4lim_(x->oo) (ln(x)')/((x^3)')`

= `4lim_(x->oo) (1/x)/(3x^2)`

= `4lim_(x->oo) (1/(3x^3))`

so on plugging the value x= oo we get

= `4(1/(3(oo)^3))`

= `0`

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