Thoroughly stumped! What is lim (n -> oo) [(log (n))^(log (2n))]/n ?

Question

I've already spent hours on this and it's driving me nuts. I've tried L'Hopital's rule and either I'm doing something wrong or it really does get messy. I've tried molding it into the form log n/n with no luck... I would be greatly appreciative of anyone who can shed some light on this problem for me. This is just a practice problem, so it's not even anything that would have a direct impact on my grade, I just want to understand it. Thanks in advance.

In response to JA: It's a limit, as the independent variable goes to infinity, of a function. In this case, the independent variable is n, and the function is [log(n) ^ log(2n)] / n. If you haven't had calculus, you won't know what I'm talking about.

P.S. An answer without any justification is not helpful at all.

Sigh... a mathematical justification, please....

OK, so I just found out that my homework assignment was wonderfully vague and didn't intend for me to formally solve it, just use intuition to basically guess. This irritates me, but c'est la vie. Thanks for all the answers.

trivialstein - Thanks for your answer. The first step you did is incorrect because the quantity (log n) is raised to (log 2n). This step would be correct if it were log (n^(log 2n)) (compared to [log n]^[log 2n]), but unfortunately it wasn't.

Answer 1

I can give you an idea as to how to simplify the equation:

log(2n) = log 2+log n

Your equation becomes:

[logn^(log2 + logn)] / n = (logn^log2) * (logn^logn) / n.

Replace logn = a, therefore: n = 10^a. Of course:

n->oo means a->oo as well.

Your equation becomes:

lim a->oo [a^log2]*[a^a] / 10^a.

This most likely will be indetermined, so you will have to use Hopital.

Remember:

d(a^a)/da :

ln (a^a) = a ln a

d(ln(a^a)) / da = d(a lna)/da = 1+lna

But, d(ln(a^a))/da = (1/a^a) *d(a^a)/da.

So: d(a^a)/da = a^a * (1+lna)

Answer 2

I tried the other day to help you and I failed miserably...Let's see if I do better this time around.

Using the property of (log a)^b = b * log (a)

You can transform (log n) ^ log (2n) into log (2n) * log (n)

Thus applying Lim (log (2n) * log (n)) / n and applying L'Hopital Rule and noting the use of the product rule...1st time through you'll get..

Lim [ 1/n * log (2n) + 1/(2n) * log (n) ] / 1..factoring the 1/n out gives you...

Lim [ Log (2n) + 1/2 log (n) ] / n. Apply L'hopitals again..

Lim [ 1/(2n) + 1/(2n)] / 1. and you have

As n ---> Infinity, Lim 1 / n = zero.

I screwed up last time I tried to help so......let me know if I did something completely mathematically wrong...I'm here to learn and spread what I know or don't know...

*edit* I did this based on the assumption that its base 10.; otherwise you have to use the properties of the above response.

Answer 3

evaluate the *series* summing words from this series. Does it converge? permit's do ratio try. we would desire to learn the restricting behaviour of the ratio between the (n + a million)th term and the nth term. as quickly as we do our simplifications, we come all the way down to: (n + a million)^2 / (2n + 2)(2n + a million) = (n + a million) / 2(2n + a million) As n techniques infinity, this techniques a million/4, it is under a million, so the series converges. What does this say relating to the series? nicely, divergence try tells us it is going to converge to 0.

Answer 4

Infinity. (log(n))^log(2n) grows faster than n.

Answer 5

its infinity

why? because my algebraic solver says so. I'm much too lazy to take it myself.
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edit:

justification? it really isn't that hard.

use l'hopital's rule

(ln(ln(x)) + ln(2)/(x*ln(x))+1/x)*ln(x)^(ln (2x))/1

lim ((ln(ln(x)) + ln(2)/(x*ln(x))+1/x)*ln(x)^(ln (2x)))
x->infinity

this is trivial. distribute if you want, but

ln(x)^(ln(2x))) ->infinity as x->infinity
ln(2)/(x*ln(x))+1/x tends to zero
ln(ln(x)) tends to infinity.

Answer 6

huh? whats this..........