solve-lim-n-n-2-ln-n-sin-1-n- Tinku Tara June 4, 2023 Limits 0 Comments FacebookTweetPin Question Number 167624 by mnjuly1970 last updated on 21/Mar/22 solveΩ=limn→∞n2.ln(n.sin(1n))=? Answered by qaz last updated on 21/Mar/22 limnn→∞2ln(nsin1n)=limnn→∞2ln(1−16n2+O(1n4))=limnn→∞2((−16n2+O(1n4))=−16 Commented by mnjuly1970 last updated on 22/Mar/22 ✓ Answered by LEKOUMA last updated on 21/Mar/22 limn→∞n2.nsin(1n)ln(n.sin(1n))n.sin(1n)limn→∞n3sin(1n)Poson,t=1n⇒n=1tsin∞,t=0limt→01t3sint=limt→0sintt3Hospitallimt→0(sint)′(t3)′=limt→0cost3t2=limt→0(cost)′(3t2)′limt→0−sint6t=−16limt→0sintt=−16limn→∞ln(n.sin(1n))n.sin(1n)=0limn→∞n2ln(n.sin(1n))=0 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-167627Next Next post: Question-167628 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.