calculus - Calculating this limit without l'Hospital/Taylor/derivatives

Let's consider

$$a_n = n^2 \log \left(\cos \frac 1n\right)$$

It's easy to calculate

$$\lim_{n\to\infty} a_n$$ 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 $-1/2.$