# I need help finding the following limit.

I cannot find the following limit: $$\lim_{x\to 0} \frac{1-\cos x \cos(2x)}{x^2} \, .$$ Please, help me.

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Apply l'Hopital's rule! – Atul Ingle Oct 15 '12 at 21:08

This can be evaluated by applying L'Hôpital's Rule twice. We have:

$$\lim_{x \to 0} \left(\frac{1-\cos x \cos(2x)}{x^2}\right) = \ldots,$$ $$= \lim_{x \to 0} \left(\frac{\sin x\cos(2x)+2\cos x\sin(2x)}{2x}\right),$$ $$= \lim_{x \to 0} \left(\frac{5\cos x \cos(2x)-4\sin x\sin(2x)}{2}\right),$$ $$= \frac{5}{2} \, .$$

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$$\cos(x) \cos(2x) = \left(1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots \right)\left(1 - \dfrac{(2x)^2}{2!} + \dfrac{(2x)^4}{4!} - \cdots \right)\\ = 1 - \dfrac{x^2 + 4x^2}{2!} + \mathcal{O}(x^4)$$ 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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Thank you but I need simpler solution. I can't use the expansion of cos(x). Do you have any idea how to reduce the problem to the limit sin(x)/x? – Adam Oct 15 '12 at 21:14
@AdamAndersson Your question never indicated any such constraint. Care to revise it? – Erick Wong Oct 15 '12 at 21:22

Let's see one line proof:

$$\lim_{x\to 0} \frac{1-\cos x \cos(2x)}{x^2}=\lim_{x\to 0}\frac{1-\cos x }{x^2}+ \lim_{x\to 0}4\cos x\frac{1-\cos 2x }{(2x)^2}=\frac{5}{2}.$$

Chris.

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Here is a proof using Tools 1-3 below and, I think, nothing else.

Tool 1: $2\cos(a)\cos(b)=\cos(a+b)+\cos(a-b)$.

Tool 2: $1-\cos(a)=2\sin^2(\frac{a}2)$.

Tool 3: Let $u(z)=\frac{\sin(z)}z$, then $u(z)\to1$ when $z\to0$. (You said you knew that.)

Using Tools 1 and 2, one gets $$\frac{1-\cos x\cos 2x}{x^2}\stackrel{\text{(Tool 1)}}{=}\frac{1-\cos 3x}{2x^2}+\frac{1-\cos x}{2x^2}\stackrel{\text{(Tool 2)}}{=}\frac{\sin^2(x/2)}{x^2}+\frac{\sin^2(3x/2)}{x^2},$$ hence $$\frac{1-\cos x\cos 2x}{x^2}=\frac{u^2(\frac{x}2)+9u^2(\frac{3x}2)}4.$$ Now, Tool 3 yields $u(\frac{x}2)\to1$ and $u(\frac{3x}2)\to1$, hence the limit is $\frac{1^2+9\cdot1^2}4=\frac52$.

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