<s>[INST] Finding the limit $ \lim_{x\to 0} \frac{1-\cos x \cos(2x)}{x^2}$ I cannot find the following limit:
$$ \lim_{x\to 0} \frac{1-\cos x \cos(2x)}{x^2} \, . $$
Please, help me.
 [/INST] $$\cos(x) \cos(2x) = \left(1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots \right)\left(1 - \dfrac{(2x)^2}{2!} + \dfrac{(2x)^4}{4!} - \cdots \right)\\ = 1 - \dfrac{x^2 + 4x^2}{2!} + \mathcal{O}(x^4)$$
Hence, $$1 - \cos(x) \cos(2x) = \dfrac52 x^2 + \mathcal{O}(x^4)$$
Can you now find the limit?
EDIT
Alternately, you could also proceed as below.
$$\cos(x) \cos(2x) = \cos(x) \left(2 \cos^2(x)-1 \right) = 2 \cos^3(x) - \cos(x)$$
Hence, $$1- \cos(x) \cos(2x) = 1 + \cos(x) - 2\cos^2(x) = (1-\cos(x))(1+2\cos(x) + 2\cos^2(x))\\ = 2 \sin^2(x/2)(1+2\cos(x) + 2\cos^2(x))$$
Hence, $$f(x) = \dfrac{1- \cos(x) \cos(2x)}{x^2} = \dfrac{2 \sin^2(x/2)(1+2\cos(x) + 2\cos^2(x))}{x^2}$$
$$\lim_{x \to 0} f(x) = \lim_{x \to 0}\dfrac{2 \sin^2(x/2)}{x^2} \times \lim_{x \to 0}(1+2\cos(x) + 2\cos^2(x)) = 2 \times \dfrac14 \times (1+2+2) = \dfrac52$$
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