If $$\lim_{x \to 0}\left\{\frac{1}{x^8}\left(1 - \cos\frac{x^2}{2} - \cos\frac{x^2}{4} + \cos\frac{x^2}{2}\cos\frac{x^2}{4}\right)\right\} = 2^{-k}$$ then the value of k is

We have to evaluate the limit

$$$L=\lim_{x\to 0}\left\{\dfrac{1}{x^{8}}\Bigl(1-\cos\dfrac{x^{2}}{2}-\cos\dfrac{x^{2}}{4}+\cos\dfrac{x^{2}}{2}\,\cos\dfrac{x^{2}}{4}\Bigr)\right\}.$$$

Whenever a limit contains trigonometric functions near zero, the easiest path is to expand every cosine in its Maclaurin (Taylor about $$0$$) series. The standard series we shall use is

$$\cos z \;=\;1-\dfrac{z^{2}}{2!}+\dfrac{z^{4}}{4!}-\dfrac{z^{6}}{6!}+\dots$$

Because the final expression will later be divided by $$x^{8}$$, terms up to the power $$x^{8}$$ must be retained. Higher-degree terms will vanish in the limit.

Let us introduce the abbreviations

$$a=\dfrac{x^{2}}{2},\qquad b=\dfrac{x^{2}}{4}.$$

Plainly $$b=\dfrac{a}{2}$$. We shall now expand each cosine.

First, for $$\cos a$$:

$$$ \begin{aligned} \cos a&=1-\dfrac{a^{2}}{2}+\dfrac{a^{4}}{24}+O(a^{6})\\ &=1-\dfrac{(x^{2}/2)^{2}}{2}+\dfrac{(x^{2}/2)^{4}}{24}+O(x^{10})\\ &=1-\dfrac{x^{4}}{8}+\dfrac{x^{8}}{384}+O(x^{10}). \end{aligned} $$$

Next, for $$\cos b$$:

$$$ \begin{aligned} \cos b&=1-\dfrac{b^{2}}{2}+\dfrac{b^{4}}{24}+O(b^{6})\\ &=1-\dfrac{(x^{2}/4)^{2}}{2}+\dfrac{(x^{2}/4)^{4}}{24}+O(x^{10})\\ &=1-\dfrac{x^{4}}{32}+\dfrac{x^{8}}{6144}+O(x^{10}). \end{aligned} $$$

The next task is to obtain the product $$\cos a\,\cos b$$. Multiplying the two truncated series and keeping all terms through $$x^{8}$$ gives

$$$ \begin{aligned} \cos a\,\cos b &=\Bigl(1-\dfrac{x^{4}}{8}+\dfrac{x^{8}}{384}\Bigr) \Bigl(1-\dfrac{x^{4}}{32}+\dfrac{x^{8}}{6144}\Bigr)+O(x^{10})\\ &=1-\dfrac{x^{4}}{32}-\dfrac{x^{4}}{8}+\dfrac{x^{8}}{256} +\dfrac{x^{8}}{6144}+\dfrac{x^{8}}{384}+O(x^{10})\\ &=1-\dfrac{5x^{4}}{32}+\dfrac{41x^{8}}{6144}+O(x^{10}). \end{aligned} $$$

Now the entire numerator in the given limit can be assembled. We write

$$$N(x)=1-\cos\dfrac{x^{2}}{2}-\cos\dfrac{x^{2}}{4}+\cos\dfrac{x^{2}}{2}\cos\dfrac{x^{2}}{4},$$$

and substitute the three expansions computed above:

$$$ \begin{aligned} N(x)=&\;1-\Bigl(1-\dfrac{x^{4}}{8}+\dfrac{x^{8}}{384}\Bigr) -\Bigl(1-\dfrac{x^{4}}{32}+\dfrac{x^{8}}{6144}\Bigr) +\Bigl(1-\dfrac{5x^{4}}{32}+\dfrac{41x^{8}}{6144}\Bigr)+O(x^{10}). \end{aligned} $$$

We carefully combine like terms.

• The constant terms: $$1-1-1+1 = 0$$.
• The $$x^{4}$$ terms: $$\dfrac{x^{4}}{8}+\dfrac{x^{4}}{32}-\dfrac{5x^{4}}{32}=0$$.
• The $$x^{8}$$ terms: $$$ -\dfrac{x^{8}}{384}-\dfrac{x^{8}}{6144}+\dfrac{41x^{8}}{6144} = -\dfrac{16x^{8}}{6144}-\dfrac{x^{8}}{6144}+\dfrac{41x^{8}}{6144} =\dfrac{24x^{8}}{6144} =\dfrac{x^{8}}{256}. $$$

Thus

$$N(x)=\dfrac{x^{8}}{256}+O(x^{10}).$$

Returning to the original limit, we divide by $$x^{8}$$:

$$$ \begin{aligned} L&=\lim_{x\to 0}\dfrac{N(x)}{x^{8}} =\lim_{x\to 0}\dfrac{\dfrac{x^{8}}{256}+O(x^{10})}{x^{8}} =\lim_{x\to 0}\left(\dfrac{1}{256}+O(x^{2})\right) =\dfrac{1}{256}. \end{aligned} $$$

We recognize that

$$\dfrac{1}{256}=2^{-8},$$

so the limit equals $$2^{-k}$$ with $$k=8$$.

So, the answer is $$8$$.