Strange l'Hospital's Rule with ln

[phillyolly]

Homework Statement

lim(x→∞)(lnx)^2/x

Homework Equations

The Attempt at a Solution

Since lim(x→∞)(lnx )^2=∞ and lim(x→∞)x=∞, the function is indeterminate, ∞/∞.

lim(x→∞)(lnx)^2/x=lim(x→∞) d/dx (lnx )^2/(d/dx x)=lim(x→∞)(2 lnx (1/x))/1

An I am stuck...

 

[stevenb]

Homework Statement

lim(x→∞)(lnx)^2/x

Homework Equations

The Attempt at a Solution

Since lim(x→∞)(lnx )^2=∞ and lim(x→∞)x=∞, the function is indeterminate, ∞/∞.

lim(x→∞)(lnx)^2/x=lim(x→∞) d/dx (lnx )^2/(d/dx x)=lim(x→∞)(2 lnx (1/x))/1

An I am stuck...

I think L'Hostpital's rule can be used very easily here. - just apply it again.

Alternatively, it's not hard to see the answer from the following expansion of logarithm.

$${\rm ln}(x)=2\sum_{i=1}^\infty {{1}\over{2i-1}}\Bigl({{x-1}\over{x+1}}\Bigr)^{2i-1}$$

With experience, you will recognize the answer to similar limits by inspection. Not many functions increase monotonically as slowly as logarithm. Granted, that's not a reliable, nor mathematically correct way to get the answer, but it helps to know the answer before you solve the problem.

 

[phillyolly]

This might be a higher-level answer. My answer should be looking much easier cause we don't work with i's and sums.

 

[stevenb]

This might be a higher-level answer. My answer should be looking much easier cause we don't work with i's and sums.

OK, don't worry about that. Just take your first answer from L'Hospital's rule and put it in a simple numerator over denominator form. Then apply the rule again.

 

[Bohrok]

(2 lnx (1/x))/1 = 2(ln x)/x

Can you finish it from there?

 

[phillyolly]

Okay, so here it is:
lim(x→∞) (2 lnx)/x=
lim(x→∞)(d/dx lnx)/(d/dx x)=
lim(x→∞)(2 (1/x))/1=
lim(x→∞)(2/x)

Should I take a derivative again, or it's enough and 2/infinity equals zero?

 

[Bohrok]

You have it, the answer is 0. You can't use l'Hopital's rule again anyway since 2/x isn't indeterminate like the other expressions you had before.

 

[phillyolly]

I see it now, thank you very much.