# Limit problem / l'hopital's rule

1. Jan 25, 2012

### gr3g1

The limit n->infinity I have to compute is:

$\frac{n\cdot \log ^{5}(n)}{n^{2}}$

Should I use L'hopital's rule? If I do, I have a problem:

First I simplify and get:

$\frac{ \log ^{5}(n)}{n}$

Taking the derivative of the top, and the bottom leads to:

$\frac{\frac{ 5\log ^{4}(n)}{n}}{1}$

At this point, we can see that the numerator is approaching 0, as n increases.

There 0/1 = 0

Is this correct?

Thanks

Last edited: Jan 25, 2012
2. Jan 25, 2012

### HallsofIvy

Why is it "obvious" (since you state it without proof) that $5log^4(n)/n$ goes to 0 when it was not obvious for $log^5(n)/n$?

What do you get if you use L'Hopital again? And Again?

3. Jan 25, 2012

### gr3g1

I just plugged in very large numbers and saw that it will approach zero. I guess I could have done that with the initial function as well.

4. Jan 25, 2012

### Dick

The conclusion is correct. But I don't see why $\frac{ \log ^{4}(n)}{n} \rightarrow 0$ is any more obvious than $\frac{ \log ^{5}(n)}{n} \rightarrow 0$. Why don't you keep applying l'Hopital until the log disappears?

5. Jan 25, 2012

### gr3g1

Since the denominator becomes one, should I just continue taking the derivative of the numerator, until log disappears?

Thanks

6. Jan 25, 2012

### Dick

You applied l'Hopital to $\frac{ \log ^{5}(n)}{n}$ correctly. The original denominator of n became 1 but now you get a new denominator of n coming from the derivative of the log.

7. Jan 25, 2012

### gr3g1

I think I see where this is going. I'll keep taking the derivative and eventually end up with a constant over n. Correct?

8. Jan 25, 2012

### Dick

Yessss.

9. Jan 25, 2012

### gr3g1

Thank you so much!