$$\lim_{x \to 0} \frac{\sin^2(\pi\cos^4 x)}{x^4}$$ is equal to:

We have to evaluate the limit

$$\lim_{x \to 0}\dfrac{\sin^{2}\!\bigl(\pi\cos^{4}x\bigr)}{x^{4}}\;.$$

Because the variable $$x$$ is tending to zero, it is natural to replace every transcendental function by its Maclaurin (Taylor) expansion around $$x=0$$. We first expand the cosine function itself.

The standard Maclaurin expansion is stated as

$$\cos x \;=\;1-\dfrac{x^{2}}{2}+\dfrac{x^{4}}{24}+O(x^{6}).$$

Now we need $$\cos^{4}x$$, so we raise the above series to the fourth power. To do this smoothly, we write

$$\cos x \;=\;1+\delta,$$

where

$$\delta \;=\;-\dfrac{x^{2}}{2}+\dfrac{x^{4}}{24}+O(x^{6}).$$

Using the binomial theorem

$$\bigl(1+\delta\bigr)^{4} \;=\;1+4\delta+6\delta^{2}+4\delta^{3}+\delta^{4},$$

and remembering that we only need terms up to $$x^{4}$$ (since the denominator already contains $$x^{4}$$), we calculate term by term.

First,

$$4\delta \;=\;4\left(-\dfrac{x^{2}}{2}+\dfrac{x^{4}}{24}\right) \;=\;-2x^{2}+\dfrac{x^{4}}{6}.$$

Next,

$$\delta^{2} \;=\;\left(-\dfrac{x^{2}}{2}\right)^{2}+O(x^{6}) \;=\;\dfrac{x^{4}}{4}+O(x^{6}),$$

and so

$$6\delta^{2} \;=\;6\cdot\dfrac{x^{4}}{4}+O(x^{6}) \;=\;\dfrac{3x^{4}}{2}+O(x^{6}).$$

The terms $$\delta^{3}$$ and $$\delta^{4}$$ start with powers $$x^{6}$$ and higher, so they are beyond the accuracy we require and can be written collectively as $$O(x^{6})$$.

Collecting the pieces, we arrive at

$$\cos^{4}x \;=\;1-2x^{2}+\left(\dfrac{1}{6}+\dfrac{3}{2}\right)x^{4}+O(x^{6}) \;=\;1-2x^{2}+\dfrac{5}{3}x^{4}+O(x^{6}).$$

Multiplying by $$\pi$$ gives

$$\pi\cos^{4}x \;=\;\pi\bigl(1-2x^{2}+\tfrac{5}{3}x^{4}+O(x^{6})\bigr) \;=\;\pi-2\pi x^{2}+\dfrac{5\pi}{3}x^{4}+O(x^{6}).$$

It is convenient to denote the small deviation from $$\pi$$ by

$$y \;=\;-2\pi x^{2}+\dfrac{5\pi}{3}x^{4}+O(x^{6}),$$

so that

$$\pi\cos^{4}x \;=\;\pi+y.$$

Now we turn to the sine term. A fundamental trigonometric identity tells us

$$\sin(\pi+y) \;=\;-\sin y.$$

Because we shall finally square the sine, the minus sign will disappear, and we may write

$$\sin^{2}\!\bigl(\pi\cos^{4}x\bigr) \;=\;\sin^{2}(\pi+y) \;=\;\sin^{2}y.$$

Next we expand the sine of a small argument. The Maclaurin expansion reads

$$\sin y \;=\;y-\dfrac{y^{3}}{6}+O(y^{5}).$$

Squaring both sides, we obtain

$$\sin^{2}y \;=\;y^{2}-\dfrac{y^{4}}{3}+O(y^{6}).$$

Because $$y$$ itself is proportional to $$x^{2}$$, the term $$y^{4}$$ is already of order $$x^{8}$$, which is far smaller than we need. Thus, for the present limit, it suffices to keep only the leading piece:

$$\sin^{2}y \;=\;y^{2}+O(x^{6}).$$

Let us compute $$y^{2}$$ up to the required accuracy:

$$y^{2} \;=\;\Bigl(-2\pi x^{2}+\dfrac{5\pi}{3}x^{4}+O(x^{6})\Bigr)^{2} \;=\;4\pi^{2}x^{4}+O(x^{6}).$$

Therefore,

$$\sin^{2}\!\bigl(\pi\cos^{4}x\bigr) \;=\;4\pi^{2}x^{4}+O(x^{6}).$$

Finally we divide this by $$x^{4}$$, exactly as dictated by the original limit:

$$\dfrac{\sin^{2}\!\bigl(\pi\cos^{4}x\bigr)}{x^{4}} \;=\;\dfrac{4\pi^{2}x^{4}+O(x^{6})}{x^{4}} \;=\;4\pi^{2}+O(x^{2}).$$

As $$x \to 0,$$ the error term $$O(x^{2})$$ vanishes, leaving

$$\lim_{x \to 0}\dfrac{\sin^{2}\!\bigl(\pi\cos^{4}x\bigr)}{x^{4}} \;=\;4\pi^{2}.$$

Hence, the correct answer is Option C.