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Answer :[cos2x^1/2]sinx= [(2cos^2-1)^1/2]sinxlet cos x = t= -[(2t^2-1)^1/2]=-sqrt2[( (2t)^2 - (sqrt2)^2 ) ^1/2 ]

The Answer is zero. Its simplified by using the " Squeezeplay theorem " or Sandwich theorem. :P ( the following steps may not explained thoroughly , but they are valid theoretically )Since, x----> Infinity,1/x ----> 0.Substitute in the equation. lim x--> x{ sin(1/x) } Since the sin f(X) is bound by +1 and -1 and oscillates about these values.. -1 <= sin (1/x) <= 1-x <= x{ sin(1/x) } <= 1Applying limits... lim x--> 0 ( -x ) <= lim x---> 0 x{sin(1/x)} <= lim x--->.0 (x) SInce the function is bounded by both values , 0 <= lim x--->0 x{sin(1/x) } <= 0 It takes the value of 0.therefore, lim x---->0 x{sin(1/x)} = 0

0 to infinity.. and If " e " itself has a range of about 2.6-3 . Accurately 2.7

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