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anonymous
 3 years ago
Can I figure out if this sequence is converging with the squeeze theorem?
Lim n>inf. (ln n)^2 /n ?
anonymous
 3 years ago
Can I figure out if this sequence is converging with the squeeze theorem? Lim n>inf. (ln n)^2 /n ?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did this \[0 \le \ln(n)^{2} \le 1/n \] solving for the limit and both side gives 0 therefore it must give 0? :/ or am I doing something wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I believe he means (ln(n))^2

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.1Yeah, ln^2(n) is >1 so you can't use that.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait I thought (ln(n))^2 meant (ln(n)) * (ln(n))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh yeah I see it it could be greater than one, stupid question :/

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.1It does. ln^2(n) is the same as (ln(n))^2. Putting the squared on the function means that. It's so you don't confuse things like sin^2(x) and sin(x^2). But anyways, what I'm saying Marc is that you can't say: ln^2(n)/n≤1/n because it isn't true.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, just noticed it, anyway thanks, basic math issues I guess ^^

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.1Yeah lol ok so your left hand side should also be something better than 0 I think, that way you're actually squeezing. ln(n)/n<ln^2(n)/n for the right side I'm still thinking haha.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well I could use l'hospital rule, I just use it so often and my answer key doesn't which gives me wrong answers so I'm afraid to use it now x.x

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.1Haha there's no "s" in L'Hopital.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yikes. Anyway thanks hehe!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0L'hopital's rule would be much easier to apply here than the squeeze theorem:\[\lim_{n \rightarrow \infty}\frac{ \ln(n)^2 }{ n }=2\lim_{n \rightarrow \infty}\frac{ \ln(n) }{ n }=2\lim_{n \rightarrow \infty}\frac{ 1 }{ n }=2*0=0\] @MarcLeclair @Kainui

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm I agree with you here, but I thought the question was specifically asking to solve this with the squeeze theorem.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well you are supposed to use squeeze theorem which he figured already, but l'hopitals would be easier if we were allowed to use it.

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.1Haha yeah for sure. Actually we can just use that simple idea to squeeze it between these two which are easy to also find the limit of with l'hopital's! haha ln(n)/n≤ln^2(n)/n≤ln^3(n)/n ?
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