Let's consider an=n2log(cos1n)
It's easy to calculate lim 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. \lim_{x\to 0}\sin x/x = 1. 2. \lim_{x\to 0}(1+ax +o(x))^{1/x} = e^a for any constant a.
From 1. we get, as x\to 0,
\frac{1-\cos x}{x^2} = \frac{1}{1+\cos x}\frac{1-\cos^2 x}{x^2} = \frac{1}{1+\cos x}\frac{\sin^2 x}{x^2} \to \frac{1}{2}\cdot 1^2 = \frac{1}{2}.
This shows \cos x = 1 - (1/2)x^2 + o(x^2). Therefore
[\cos(1/n)]^{n^2} = [1+(-1/2)/n^2 + o(1/n^2)]^{n^2} \to e^{-1/2},
where we have used 2. above. Now apply \ln to see the desired limit is -1/2.
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