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Question Number 140046 by mnjuly1970 last updated on 03/May/21
...........nice.......calculus(I)........Θ:=limx→π4(tan(x))tan(2x)=?...............................
Commented by mnjuly1970 last updated on 03/May/21
thanksalot...
Answered by Dwaipayan Shikari last updated on 03/May/21
limz→0tan(π2+2z)log(tan(π4+z))=log(y)−cot(2z)log(1+tanz1−tanz)=log(y)∼−cot(2z)log(1+z)+cot(2z)log(1−z)=log(y)≈−12zz+−z2z=log(y)⇒log(y)=−1⇒y=1e
Answered by mnjuly1970 last updated on 04/May/21
solution.....tan(x):=y⇒{y→1x→π4Θ:=limy→1(1−(1−y))2y(1−y)(1+y):=1−y=tlimt→0(1−t)2(1−t)t(2−t)=limt→0{(1−t)2t}1−t2−t:=(e−2)12=e−1=1e......✓✓.........Θ:=1e........
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