# Homework Help: Strange l'Hospital's Rule with ln

1. Jul 10, 2010

### phillyolly

1. The problem statement, all variables and given/known data

lim(x→∞)(lnx)^2/x

2. Relevant equations

3. 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...

2. Jul 10, 2010

### stevenb

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.

3. Jul 10, 2010

### 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.

4. Jul 10, 2010

### stevenb

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.

5. Jul 10, 2010

### Bohrok

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

Can you finish it from there?

6. Jul 10, 2010

### 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?

7. Jul 10, 2010

### 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.

8. Jul 10, 2010

### phillyolly

I see it now, thank you very much.