1. We need to evaluate

$$L=\lim_{x\to 0}\frac{\sin\left(\pi\cos^2 x\right)}{x^2}.$$

2. First, simplify the angle inside the sine using

$$\cos^2 x = 1-\sin^2 x.$$

So,

$$\pi\cos^2 x = \pi(1-\sin^2 x)=\pi-\pi\sin^2 x.$$

Hence,

$$\sin(\pi\cos^2 x)=\sin(\pi-\pi\sin^2 x).$$

Using the identity

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

we get

$$\sin(\pi\cos^2 x)=\sin(\pi\sin^2 x).$$

Therefore,

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

3. Now use the standard limit

$$\lim_{u\to 0}\frac{\sin u}{u}=1.$$

Write

$$L=\lim_{x\to 0}\left(\frac{\sin(\pi\sin^2 x)}{\pi\sin^2 x}\right)\left(\frac{\pi\sin^2 x}{x^2}\right).$$

As $x\to 0$, we have $\pi\sin^2 x\to 0$, so

$$\frac{\sin(\pi\sin^2 x)}{\pi\sin^2 x}\to 1.$$

Also,

$$\frac{\sin^2 x}{x^2}=\left(\frac{\sin x}{x}\right)^2 \to 1.$$

Thus,

$$L=1\cdot \pi \cdot 1=\pi.$$

4. Therefore the limit is

$$\boxed{\pi}.$$

5. Option check:

• A: $-\pi$ ❌
• B: $\pi$ ✅
• C: $\dfrac{\pi}{2}$ ❌
• D: $1$ ❌