Limit problem / l'hopital's rule

[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

 

[HallsofIvy]

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

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?

 

[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.

 

[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?

 

[gr3g1]

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

Thanks

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?

 

[Dick]

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

Thanks

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.

 

[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?

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.

 

[Dick]

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

Yessss.

 

[gr3g1]

Thank you so much!