Let's consider an=n2log(cos1n)
It's easy to calculate limn→∞an
by using l'Hospital/Taylor. But how to do it without using anything that involves derivatives? (Pure sequences!)
Answer
I'll use two facts 1. limx→0sinx/x=1. 2. limx→0(1+ax+o(x))1/x=ea for any constant a.
From 1. we get, as x→0,
1−cosxx2=11+cosx1−cos2xx2=11+cosxsin2xx2→12⋅12=12.
This shows cosx=1−(1/2)x2+o(x2). Therefore
[cos(1/n)]n2=[1+(−1/2)/n2+o(1/n2)]n2→e−1/2,
where we have used 2. above. Now apply ln to see the desired limit is −1/2.
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