Question Number 103749 by bobhans last updated on 17/Jul/20

lim_(n→∞)  2^n  sin ((π/2^n )) = ?

Answered by bramlex last updated on 17/Jul/20

set x = (π/2^n ) ⇒2^n  = (π/x) ;x→0  lim_(x→0)  (π/x).sin x = π lim_(x→0)  ((sin x)/x) = π .  □

Answered by Sobir last updated on 17/Jul/20

the  ansver is   π

Answered by Dwaipayan Shikari last updated on 17/Jul/20

lim_(n→∞) π((sin(π/2^n ))/(π/2^n ))=π