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MarcLeclair
Can I figure out if this sequence is converging with the squeeze theorem? Lim n-->inf. (ln n)^2 /n ?
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
I believe he means (ln(n))^2
Yeah, ln^2(n) is >1 so you can't use that.
Wait I thought (ln(n))^2 meant (ln(n)) * (ln(n))
Oh yeah I see it it could be greater than one, stupid question :/
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.
Yeah, just noticed it, anyway thanks, basic math issues I guess ^^
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.
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
Haha there's no "s" in L'Hopital.
Yikes. Anyway thanks hehe!!
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
Hmm I agree with you here, but I thought the question was specifically asking to solve this with the squeeze theorem.
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.
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 ?