Applying L'Hôpital's rule to lim x→∞ (ln x)²/x

[shocklightnin]

Homework Statement

This was a question from our lecture notes, just not sure how the prof arrived at the answer.

lim x->infinity (lnx)^2/x

Homework Equations

lim x->infinity (lnx)^2/x
lim x->infinity 2lnx/x

The Attempt at a Solution

so both the numerator and denominator are going towards infinity, and by L'H it the lim x->infinity 2lnx/x
so this means that the numerator is 'growing' faster than the denominator, a constant x? also, how does one arrive at the conclusion that the limit is 0?

 

[tazzzdo]

(lnx)^2/x = (2/x)(ln x)

derivative of ln x = 1/x

Go from there. I'm drunk.

 

[RoshanBBQ]

Homework Statement

This was a question from our lecture notes, just not sure how the prof arrived at the answer.

lim x->infinity (lnx)^2/x

Homework Equations

lim x->infinity (lnx)^2/x
lim x->infinity 2lnx/x

The Attempt at a Solution

so both the numerator and denominator are going towards infinity, and by L'H it the lim x->infinity 2lnx/x
so this means that the numerator is 'growing' faster than the denominator, a constant x? also, how does one arrive at the conclusion that the limit is 0?

Just apply L'Hospital's rule again since you are still in an indeterminate inf/inf:
$$\frac{2}{x}$$

It should now be pretty sensible that it approaches 0 as x approaches infinity.

 

[shocklightnin]

RoshanBBQ, thanks! Completely slipped my mind that sometimes we have to apply L'H more than once.