calculate-lim-x-0-ln-1-x-sinx-ln-1-sin-2x-x-2- Tinku Tara June 4, 2023 Limits 0 Comments FacebookTweetPin Question Number 62200 by maxmathsup by imad last updated on 17/Jun/19 calculatelimx→0ln(1+x+sinx)−ln(1+sin(2x))x2 Commented by maxmathsup by imad last updated on 18/Jun/19 letusehospitaltheoremletu(x)=ln(1+x+sinx)−ln(1+sin(2x)⇒u′(x)=1+cosx1+x+sinx−2cosx1+sin(2x)u″(x)=−sinx(1+x+sinx)−(1+cosx)2(1+x+sinx)2−2−sinx(1+sin(2x))−cosx(2cosx(2x))(1+sin(2x))2⇒limx→0u(2)(x)=−41−2−21=−4+4=0v(x)=x2⇒v′(x)=2x⇒v(2)(x)=2⇒limx→0ln(1+x+sinx)−ln(1+sin(2x))x2=limx→0u(2)(x)v(2)(x)=0 Answered by tanmay last updated on 17/Jun/19 limx→0ln(1+x+sinx1+sin2x)x2limx→0ln(1+1+x+sinx1+sin2x−1)(1+x+sinx1+sin2x−1)×x+sinx−sin2x1+sin2xx2limt→0ln(1+t)t×limx→011+sin2x×limx→0x+sinx−sin2xx21×1×limx→01+cosx−2cos2x2x(00)=limx→0−sinx+4sin2x2=0 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: let-f-x-e-1-x-determine-f-n-by-relation-of-recurrence-Next Next post: study-the-convergence-of-n-1-n-1-n-nln-n-1- Leave a Reply Cancel replyYour email address will not be published.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.