Evaluating $\lim\limits_{x\to 0}\!\left[\! \frac{1 - (\cos(2x) \cos(4x) \cos(6x) \cos(8x) \cos(10x))^3}{5 \tan^2x}\!\right]$

Question

Following question is given in my book:

Let $f(x) = \cos(2x) \cos(4x) \cos(6x) \cos(8x) \cos(10x)$ and $M = \lim\limits_{x\to 0 } \left[\dfrac{1 - f(x)^3}{5 \tan^2x}\right]$, where $M$ is finite.
Then what is the value of $\sqrt{M - 2} + 1$.

My work: \begin{align} M &= \lim_{x\to 0}\left[\frac{1 - f(x)^3}{5 \tan^2x}\right]\\ & = \lim_{x\to 0}\left[\frac{(1 - f(x)) (1 + f(x) + f(x)^2)}{5 \tan^2(x)}\right]\\& =\frac 15 \lim_{x\to 0}\left[\frac{(1 - f(x)) (1 + f(x) + f(x)^2)}{x^2}\right]\cdot \lim_{x\to 0} \frac{x^2}{\tan^2x} \\ & = \frac{3}{5} \lim_{x\to 0}\left[\frac{1 - f(x)}{x^2}\right]\tag{$*$}\end{align}

$(*)$ $1 + f(x) + f(x)^2$ evaluates to $3$ at $x = 0$.

Maybe we can use double angle formula of cosine but I'm not quite sure how to do that in the best possible manner. I also thought of converting cosine in its exponential form, but I couldn't continue from that either.

Answer 1

My approach would be to use a series expansion of $f$ about $x = 0$: note that $$\cos 2kx = 1 - \frac{(2kx)^2}{2!} + O(x^4),$$ hence $$\begin{align} f(x) &= \prod_{k=1}^5 \cos 2kx \\ &= \prod_{k=1}^5 \left(1 - 2 k^2 x^2 \right) + O(x^4) \\ &= 1 - 2(1 + 4 + 9 + 16 + 25)x^2 + O(x^4) \\ &= 1 - 110x^2 + O(x^4). \end{align}$$ This immediately yields $$\lim_{x \to 0} \frac{1 - f(x)}{x^2} = 110.$$

Answer 2

We have that $\cos 2\theta=1-2\sin^2 \theta$ then

$$f(x) = (1-2\sin^2 x)(1-2\sin^2 (2x))(1-2\sin^2 (3x))(1-2\sin^2 (4x))(1-2\sin^2(5x))$$

and

$$1-f(x)=2\sin^2 x+2\sin^2 (2x)+2\sin^2 (3x)+2\sin^2 (4x)+2\sin^2 (5x)+g(x)$$

thus we can proceed without Taylor's expansion using that $\frac{\sin x}x\to 1$ as $x \to 0$ indeed

$$\frac{1 - f(x)}{5 \tan^2x}=\frac 25 \frac{x^2}{\tan^2 x}\left(\frac{\sin^2 x}{x^2}+4\frac{\sin^2 (2x)}{(2x)^2}+9\frac{\sin^2 (3x)}{(3x)^2}+16\frac{\sin^2 (4x)}{(4x)^2}+25\frac{\sin^2 (5x)}{(5x)^2}+\frac{g(x)}{x^2}\right)\to\frac25\cdot 1\left(1+4+9+16+25+0\right)=22$$

since $g(x)$ is the sum of terms such $g_i(x)=2^k\;\overbrace{\sin^2(mx)\cdot \ldots \cdot \sin^2(nx)}^{\text{k terms}}$ with $k\ge 2$ such that

$$\frac{g_i(x)}{x^2}=2^k\frac{\sin^2(mx)\cdot \ldots \cdot \sin^2(nx)}{x^2}=2^km^2\frac{\sin^2(mx)}{(mx)^2}\cdot \ldots \cdot \sin^2(nx) \to 2^km^2\cdot 0=0$$

Answer 3

There is another possible trick using the transformation of the product to sum and multiple angle formulae.

If $$f(x)= \cos (2 x) \cos (4 x) \cos (6 x) \cos (8 x) \cos (10 x)$$then $$A=16f(x)=3 \cos (2 x)+3 \cos (6 x)+3 \cos (10 x)+2 \cos (14 x)+$$ $$2 \cos (18x)+\cos (22 x)+\cos (26 x)+\cos (30 x)$$

So, using Taylor, the constant term is $$3+3+3+2+2+1+1+1=16$$ and twice the term in $x^2$ is $$3\times 2^2+3\times 6^2+3 \times10^2+2\times 14^2+2\times 18^2+22^2+26^2+30^2=3520$$ So $$f(x)=\frac {1}{16}-\frac 1 2 \times \frac {3520}{16} x^2=1-110 x^2$$

Answer 4

To evaluate the limit of $$\frac{1-\cos 2x\cos 4x\dots\cos 10x}{x^2}$$ let us note that numerator is of the form $1-abc\dots$ where each of $a, b, c, \dots $ tends to $1$. Further limit of each of the fractions $(1-a)/x^2,(1-b)/x^2,\dots$ exists.

We can then split the numerator as $$1-abc\dots=1-a+a(1-bc\dots)$$ and thereby the original fraction gets split into two terms. Since $a\to 1$ we can see that the desired limit is equal to the limit of $$\frac{1-a}{x^2}+\frac{1-bc\dots}{x^2}$$ Applying same technique to second fraction above and continuing the process we see that desired limit equals the limit of $$\frac {1-a}{x^2}+\frac{1-b}{x^2}+\frac{1-c}{x^2}+\dots$$ Thus the limit of the fraction at the beginning of this answer is equal to the limit of $$\sum_{k=1}^5\frac{1-\cos 2kx}{x^2}$$ ie $\sum_{k=1}^52k^2=110$. The limit of the expression in question is thus $110(3/5)=66$.

The technique above is based on rules of limits and has been used elsewhere on this site to handle a complicated limit.