# `lim_(x->oo) (ln(x))^2/(x)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply,...

`lim_(x->oo) (ln(x))^2/(x)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to evaluate the limit, such that:

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

Since the limit is indeterminate `oo/oo` , you may use l'Hospital's rule:

`lim_(x->oo) ((ln^2 x)')/(x') = lim_(x->oo) ((1/x)*(2lnx))/1 = 2lim_(x->oo) (lnx)/x = oo/oo`

You may use again l'Hospital's rule:

`2lim_(x->oo) (lnx)/x = 2lim_(x->oo) (1/x)/1 = 2*1/oo = 2*0 = 0`

Hence, evaluating the limit using l'Hospital's rule twice, yields `lim_(x->oo) (ln^2 x)/x= 0.`

Educator Approved

scisser | (Level 3) Honors

Posted on

Plugging in `oo ` everywhere there is x, you get `oo/oo `

Apply LH's rule

`lim_(x-gtoo)(((2lnx)/x)/1)`

Simplify,

`lim_(x-gtoo)((2lnx)/x)`

Plug in `oo ` again, and you get `oo/oo`

Apply LH's rule again

`lim_(x-gtoo)(2/x)`

2 divided by a really big number is just 0.

`=0`