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

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MarcLeclairBest 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
 11 months ago

MarcLeclairBest ResponseYou've already chosen the best response.0
ln(n)^2/n * sorry
 11 months ago

genius12Best ResponseYou've already chosen the best response.0
I believe he means (ln(n))^2
 11 months ago

KainuiBest ResponseYou've already chosen the best response.1
Yeah, ln^2(n) is >1 so you can't use that.
 11 months ago

MarcLeclairBest ResponseYou've already chosen the best response.0
Wait I thought (ln(n))^2 meant (ln(n)) * (ln(n))
 11 months ago

MarcLeclairBest ResponseYou've already chosen the best response.0
Oh yeah I see it it could be greater than one, stupid question :/
 11 months ago

KainuiBest 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.
 11 months ago

MarcLeclairBest ResponseYou've already chosen the best response.0
Yeah, just noticed it, anyway thanks, basic math issues I guess ^^
 11 months ago

KainuiBest 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.
 11 months ago

MarcLeclairBest 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
 11 months ago

KainuiBest ResponseYou've already chosen the best response.1
Haha there's no "s" in L'Hopital.
 11 months ago

MarcLeclairBest ResponseYou've already chosen the best response.0
Yikes. Anyway thanks hehe!!
 11 months ago

genius12Best 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
 11 months ago

KainuiBest 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.
 11 months ago

genius12Best 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.
 11 months ago

KainuiBest 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 ?
 11 months ago
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