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solve-lim-n-n-2-ln-n-sin-1-n-




Question Number 167624 by mnjuly1970 last updated on 21/Mar/22
    solve    Ω =lim_( n→∞) n^( 2) . ln( n . sin((1/n)))=?
solveΩ=limnn2.ln(n.sin(1n))=?
Answered by qaz last updated on 21/Mar/22
lim_(n→∞) n^2 ln(nsin (1/n))  =lim_(n→∞) n^2 ln(1−(1/(6n^2 ))+O((1/n^4 )))  =lim_(n→∞) n^2 ((−(1/(6n^2 ))+O((1/n^4 )))  =−(1/6)
limnn2ln(nsin1n)=limnn2ln(116n2+O(1n4))=limnn2((16n2+O(1n4))=16
Commented by mnjuly1970 last updated on 22/Mar/22
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Answered by LEKOUMA last updated on 21/Mar/22
lim_(n→∞) n^2 .nsin ((1/n))((ln (n.sin ((1/n))))/(n.sin ((1/n))))  lim_(n→∞) n^3 sin((1/n))  Poson,  t=(1/n) ⇒ n=(1/t)  si n ∞, t=0  lim_(t→0) (1/t^3 )sin t=lim_(t→0) ((sin t)/t^3 )  Hospital  lim_(t→0) (((sin t)′)/((t^3 ) ′))=lim_(t→0) ((cos t)/(3t^2 ))=lim_(t→0) (((cos t) ′)/((3t^2 )^′ ))  lim_(t→0) ((−sin t)/(6t))=−(1/6)lim_(t→0) ((sin t)/t)=−(1/6)  lim_(n→ ∞) ((ln (n.sin ((1/n))))/(n.sin ((1/n))))=0  lim_(n→∞) n^2 ln (n.sin((1/n)))=0
limnn2.nsin(1n)ln(n.sin(1n))n.sin(1n)limnn3sin(1n)Poson,t=1nn=1tsin,t=0limt01t3sint=limt0sintt3Hospitallimt0(sint)(t3)=limt0cost3t2=limt0(cost)(3t2)limt0sint6t=16limt0sintt=16limnln(n.sin(1n))n.sin(1n)=0limnn2ln(n.sin(1n))=0

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