A limit involving $\cos x$ and $x^2$

Question

The question is finding the value of $L = \lim_{x \to 0}\frac{1 - \cos x \cos 2x \cos 3x}{x^2}$ if it exists . I've found the answer using taylor's formula but I'm looking for other solutions maybe using trigonometric identities .

Answer 1

You can use the formulae

$$\cos A\cos B=\frac 12 \cos(A+B)+\frac {1}{2} \cos(A-B).$$

then:

$$ \begin{align}\cos(x)\cos(2x)\cos(3x)&=\cos(x)(\cos(5x)+\sin(2x)\sin(3x))\\&=\cos(x)-\sin(x)\sin(5x)-\cos(x)\sin(2x)\sin(3x).\end{align}$$

and using the remarkable limit $\lim_{x\to0} \frac {\sin(x)}x=1.$

You get as JackyChong $18-5-6=7.$

Or:

You can directly apply L'Hospital rule and get that:

$$ \begin{align} \\&\lim_{x\to 0} \frac {1-\cos(x)\cos(2x)\cos(3x)}{x^2} \\&=\lim_{x\to0}\frac {1}{2} \frac {\cos(2x)\cos(3x)\sin(x)+2\cos(x)\cos(3x)\sin(2x)+3\cos(x)\cos(2x)\sin(3x)}{x} \\&=\frac{1+4+9}{2}=\frac {14}2=7. \end{align}$$

Answer 2

Observe you have \begin{align} \cos x \cos 2x \cos 3x= \cos x(\cos 5x+\sin 2x \sin 3x) \end{align} and \begin{align} \cos x \cos 5x =\cos 6x+ \sin x\sin 5x \end{align} which means \begin{align} \frac{1-\cos x \cos 2x \cos 3x}{x^2} =&\ \frac{1-\cos 6x}{x^2}-\frac{\sin x\sin 5x}{x^2}-\frac{\cos x \sin 2x\sin 3x}{x^2}\\ =&\ 18\frac{\sin^2 3x}{(3x)^2}-5\frac{\sin x\sin 5x}{x(5x)}-6\frac{\cos x \sin 2x\sin 3x}{(2x)(3x)} \end{align} In the limit, we have \begin{align} 18-5 - 6= 7 \end{align}

Answer 3

Note that by product to sum formula

$$\cos x \cos 3x = \frac12 (\cos 2x+\cos 4x)\\\implies \cos x \cos 2x \cos 3x = \frac12 (\cos^2 2x+\cos 2x \cos 4x)=\frac12(2\cos^3 2x+\cos^2 2x - \cos 2x )$$

then

$$\frac{1 - \cos x \cos 2x \cos 3x}{x^2}= \frac{1 - \cos^3 2x-\frac12\cos^2 2x + \frac12\cos 2x }{x^2}=\\\frac{\frac12(1-\cos 2x)(2\cos^2 2x+3\cos 2x+2)}{x^2}=\\2\frac{1-\cos 2x}{4x^2}(2\cos^2 2x+3\cos 2x+2)\to 1 \cdot 7=7$$

Answer 4

Hint: your numerator can be written as $$\sin ^2(x) (3 \cos (2 x)+\cos (4 x)+3)$$

Answer 5

Alternatively: $$\cos 3x=\cos (2x+x)=\cos 2x\cos x-\sin 2x\sin x.$$ $$L = \lim_{x \to 0}\frac{1 - \cos x \cos 2x \cos 3x}{x^2}=\lim_{x \to 0}\frac{1 - \cos x \cos 2x (\cos 2x\cos x-\sin 2x\sin x)}{x^2}=\\ \lim_{x \to 0}\frac{1 - \cos^2 x(1-\sin ^22x)+\cos x\cos 2x\sin 2x\sin x}{x^2}=\\ \lim_{x \to 0}\frac{\sin^2x + \cos^2x\sin ^22x+\cos x\cos 2x\sin 2x\sin x}{x^2}=\\ \lim_{x\to 0} \frac{\sin^2x}{x^2}+\lim_{x\to 0} \frac{4\sin^22x}{(2x)^2}+\lim_{x\to 0} \frac{2\sin 2x\sin x}{2x^2}=1+4+2=7.$$

Answer 6

An answer not using trig identities: rewrite the function as $$ \begin{align} &\frac{1}{x^2}\left[1-\left(1-x^2\frac{1-\cos x}{x^2}\right)\left(1-4x^2\frac{1-\cos 2x}{4x^2}\right)\left(1-9x^2\frac{1-\cos 3x}{9x^2}\right)\right]=\\ &\quad=\frac{1}{x^2}[1-(1-x^2f(x))(1-4x^2f(2x))(1-9x^2f(3x))]=\\ &\quad=f(x)+4f(2x)+9f(3x)-x^2[4f(x)f(2x)+9f(x)f(3x)+36f(2x)f(3x)]+36x^4f(x)f(2x)f(3x) \end{align} $$ with $$ f(x)=\frac{1-\cos x}{x^2} $$ In the limit $x\to0$, only the first three terms survive and they give $$ \frac{1}{2}(1+4+9)=7 $$

Answer 7

Hint:

$$\displaystyle \frac{1-\cos(x)\cos(2x)\cos(3x)}{x^2} = \frac{1-\cos(x)}{x^2} + \frac{\cos(x) - \cos(x)\cos(2x)}{x^2} + \frac{\cos(x)\cos(2x) - \cos(x)\cos(2x)\cos(3x)}{x^2},$$ and now use the limit $$\displaystyle \lim_{x \to 0} \frac{1-\cos(x)}{x^2} = \frac{1}{2} $$