limx→∞[cos(2π(xx+1))α]x2
where α∈Q
Try: l=limx→∞[cos(2π(1−1x+1))]x2
Put 1x+1=t. Then limit convert into l=limt→0(cos(2πt)α)1t−1
ln(l)=limt→1(1t−1)ln(cos(2πt)α)
Could some Help me to solve it, Thanks
Answer
Hint:
Put 1/x=t to find
limt→0(cos2πt+1)1/t2
=limt→0(cos(2π−2πt+1))1/t2 as cos(2π−x)=cosx
limt→0(cos2πtt+1)1/t2
=[limt→0(1−2sin2πtt+1)−12sin2πtt+1]−2limt→0(2sinπtt+1t)2
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