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

We want to evaluate the limit $$\displaystyle\lim_{x\to 0}\frac{\sin(\pi\cos^2 x)}{x^2}\;.$$

First, we expand $$\cos x$$ about $$x=0$$. Using the Maclaurin series, we know

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

Now we square this expression because we need $$\cos^2 x$$. Let $$a = -\dfrac{x^2}{2}+\dfrac{x^4}{24}+O(x^6)$$ so that $$\cos x = 1+a$$. Then

$$\cos^2 x \;=\;(1+a)^2 \;=\;1+2a+a^2.$$

Substituting $$a$$ and keeping terms up to $$x^4$$, we obtain

$$\begin{aligned} 2a &= 2\left(-\frac{x^2}{2}+\frac{x^4}{24}\right) \;=\; -x^2+\frac{x^4}{12},\\[4pt] a^2 &= \left(-\frac{x^2}{2}+\frac{x^4}{24}\right)^2 \;=\;\frac{x^4}{4}+O(x^6). \end{aligned}$$

Adding these pieces, we have

$$\cos^2 x \;=\;1 - x^2 + \frac{x^4}{12} + \frac{x^4}{4} + O(x^6) \;=\;1 - x^2 + \left(\frac{1}{12}+\frac{3}{12}\right)x^4 + O(x^6) \;=\;1 - x^2 + \frac{x^4}{3} + O(x^6).$$

Next, multiply by $$\pi$$ because our argument of the sine function is $$\pi\cos^2 x$$. Thus

$$\pi\cos^2 x \;=\;\pi\Bigl(1 - x^2 + \frac{x^4}{3} + O(x^6)\Bigr) \;=\;\pi - \pi x^2 + \frac{\pi x^4}{3} + O(x^6).$$

It will be convenient to separate this as

$$\pi\cos^2 x \;=\;\pi - \Delta,$$

where we define

$$\Delta \;=\;\pi x^2 - \frac{\pi x^4}{3} + O(x^6).$$

Notice that $$\Delta \to 0$$ as $$x \to 0$$. Now we use the well-known trigonometric identity

$$\sin(\pi - \theta) = \sin\theta.$$

With $$\theta=\Delta$$, this gives

$$\sin(\pi\cos^2 x) \;=\;\sin\!\bigl(\pi - \Delta\bigr) \;=\;\sin\Delta.$$

Because $$\Delta$$ is small, we expand $$\sin\Delta$$ using its Maclaurin series. Stating the formula:

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

Setting $$y=\Delta$$, we get

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

Therefore

$$\sin(\pi\cos^2 x) \;=\;\Delta - \frac{\Delta^3}{6} + O(\Delta^5).$$

But $$\Delta$$ itself is already of order $$x^2$$, so $$\Delta^3$$ is of order $$x^6$$, which will vanish when divided by $$x^2$$ in the limit. Hence we need only the leading term:

$$\sin(\pi\cos^2 x) \;=\;\Delta + O(x^6) \;=\;\pi x^2 - \frac{\pi x^4}{3} + O(x^6).$$

Now substitute this into the original fraction:

$$\frac{\sin(\pi\cos^2 x)}{x^2} \;=\;\frac{\pi x^2 - \frac{\pi x^4}{3} + O(x^6)}{x^2} \;=\;\pi - \frac{\pi x^2}{3} + O(x^4).$$

Finally, let $$x \to 0$$. All terms containing positive powers of $$x$$ disappear, leaving

$$\displaystyle\lim_{x\to 0}\frac{\sin(\pi\cos^2 x)}{x^2} \;=\;\pi.$$

Hence, the correct answer is Option B.