lim-x-pi-3-cos-x-sin-pi-6-pi-6-x-2- Tinku Tara June 4, 2023 Limits 0 Comments FacebookTweetPin Question Number 55592 by gunawan last updated on 27/Feb/19 limx→π/3cosx−sinπ6π6−x2=.. Commented by maxmathsup by imad last updated on 27/Feb/19 letA(x)=cosx−sin(π6)π6−x2⇒A(x)=6cosx−12π−3x=32cosx−13(π3−x)=1−2cosxx−π3changementx−π3=tgivelimx→π3A(x)=limt→01−2cos(t+π3)t=limt→01−2{12cost−32sint)}t=limt→01−cost+3sintt=limt→01−costt+limt→03sinttbutcost∼1−t22⇒1−costt∼t2→0(t→0)alsosint∼t⇒limt→0sintt=1⇒limx→π3A(x)=3. Answered by kaivan.ahmadi last updated on 27/Feb/19 hopitallimx→π3−sinx−12=limx→π32sinx=2sinπ3=2×32=3 Answered by math1967 last updated on 27/Feb/19 limx→π3cosx−cos(π2−π6)π6−x2limx→π32sin(π6−x2)sin(x2+π6)π6−x2★2×1×sin(π6+π6)=2×1×32=3★lim(π6−x2)→0sin(π6−x2)(π6−x2)=1∵x→π3∴x2→π6∴(π6−x2)→0 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-186662Next Next post: a-b-gt-0-a-b-2-Prove-that-a-2b-b-2a-a-b-2-2-2- Leave a Reply Cancel replyYour email address will not be published.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.