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MarcLeclair
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Can I figure out if this sequence is converging with the squeeze theorem?
Lim n>inf. (ln n)^2 /n ?
 one year ago
 one year ago
MarcLeclair Group Title
Can I figure out if this sequence is converging with the squeeze theorem? Lim n>inf. (ln n)^2 /n ?
 one year ago
 one year ago

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MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
I 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
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
ln(n)^2/n * sorry
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
I believe he means (ln(n))^2
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
Not ln(n)^2
 one year ago

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

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
Wait I thought (ln(n))^2 meant (ln(n)) * (ln(n))
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
Oh yeah I see it it could be greater than one, stupid question :/
 one year ago

Kainui Group TitleBest ResponseYou've already chosen the best response.1
It 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.
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
Yeah, just noticed it, anyway thanks, basic math issues I guess ^^
 one year ago

Kainui Group TitleBest ResponseYou've already chosen the best response.1
Yeah 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.
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
Well 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
 one year ago

Kainui Group TitleBest ResponseYou've already chosen the best response.1
Haha there's no "s" in L'Hopital.
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
Yikes. Anyway thanks hehe!!
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
L'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
 one year ago

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

genius12 Group TitleBest ResponseYou've already chosen the best response.0
Well you are supposed to use squeeze theorem which he figured already, but l'hopitals would be easier if we were allowed to use it.
 one year ago

Kainui Group TitleBest ResponseYou've already chosen the best response.1
Haha 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 ?
 one year ago
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