Help with understanding of L'Hospitals Rule

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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?
 
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(lnx)^2/x = (2/x)(ln x)

derivative of ln x = 1/x

Go from there. I'm drunk.
 
shocklightnin said:

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.
 
RoshanBBQ, thanks! Completely slipped my mind that sometimes we have to apply L'H more than once.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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