Question-173499 Tinku Tara June 4, 2023 Limits 0 Comments FacebookTweetPin Question Number 173499 by cortano1 last updated on 12/Jul/22 Commented by blackmamba last updated on 12/Jul/22 x=1+rL=limr→07+(1+r)33−3+(1+r)2r=limr→08+(3r+3r2+r3)3−4+(2r+r2)r=limr→021+(3r+3r2+r38)3−21+(2r+r24)r=limr→02(1+3r+3r2+r324)−2(1+(2r+r28))r=limr→03+3r+r212−2+r4=−14 Commented by Tawa11 last updated on 13/Jul/22 Greatsir Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-42422Next Next post: Bonus-du-Mardi-12-07-2022-I-0-2-sin-2-x-1-cos-2-x-dx-J-0-2-dx-1-cos-2-x-I-0-2-2-1-cos-2-x-1-cos-2-x-dx- Leave a Reply Cancel replyYour email address will not be published.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.