Limit of Logarithmic Function as n Goes to Infinity

[EV33]

Homework Statement

Does anyone have any idea how to take the limit of log(n)/sqrt(n) as n goes to infinity?

Homework Equations

The Attempt at a Solution

I know it is zero by my good old calculator but I am not aware of any theorems that I could use to solve this. If anyone could give me a hint on how to start that would be very helpful.

Thank you.

 

[malicx]

Homework Statement

Does anyone have any idea how to take the limit of log(n)/sqrt(n) as n goes to infinity?

Homework Equations

The Attempt at a Solution

I know it is zero by my good old calculator but I am not aware of any theorems that I could use to solve this. If anyone could give me a hint on how to start that would be very helpful.

Thank you.

Have you learned about L'Hopital's rules for finding limits of indeterminate forms? (in this case, infty/infty)

 

[╔(σ_σ)╝]

L'hospitals ?!

 

[Dick]

L'hospitals ?!

It used to be l'Hospital. French spelling has changed. Probably ought have the circumflex over the o though.

 

[EV33]

No we haven't learned that. We are trying to prove that this limit is zero not evaluate it!

 

[malicx]

No we haven't learned that. We are trying to prove that this limit is zero not evaluate it!

I guess you could exponentiate your sequence, show that it goes to one, then take the log at the end.

 

[Dick]

No we haven't learned that. We are trying to prove that this limit is zero not evaluate it!

If you've proved the limit is zero, then you have evaluated it. If you haven't done l'Hopital yet then try malicx idea of looking at lim log(e^n)/sqrt(e^n). Apply a 'ratio test'. Show it's bounded by a geometric sequence r^n with r<1.

 

[EV33]

Thank you Dick. But are you basically doing substitution there because evaluating that limit with some random e^x in there doesn't make sense with anything we have done in our course so far.

 

[Dick]

Thank you Dick. But are you basically doing substitution there because evaluating that limit with some random e^x in there doesn't make sense with anything we have done in our course so far.

If lim f(n) as n->infinity has a limit, then that limit is the same as the limit of f(e^n) as n->infinity. Because as n->infinity, e^n also goes to infinity.

 

[malicx]

Thank you Dick. But are you basically doing substitution there because evaluating that limit with some random e^x in there doesn't make sense with anything we have done in our course so far.

Well... lim log(n)/sqrt(n) = lim log(e^n)/sqrt(e^n), since in either case, e^n and n both go to infinity (one just does it faster!)

There are a lot of sequence theorems involved with sequences, functions, and combinations of both but I don't know which you have learned or are expected to know.

So, given that, can you squeeze your sequence between another couple sequences that have the same limit?

 

[Dickfore]

Maybe you can something about the limit of the function:

<br /> x \, \ln{x}<br />

when x \rightarrow 0.