Let $$[x]$$ denote the greatest integer less than or equal to x. Then: $$\lim_{x \to 0} \frac{\tan(\pi \sin^2 x) + (|x| - \sin(x[x]))^2}{x^2}$$

We have to find the value of

$$L=\lim_{x\to 0}\frac{\tan\!\bigl(\pi\sin^{2}x\bigr)\;+\;\bigl(|x|-\sin(x[x])\bigr)^{2}}{x^{2}}.$$

The limit can exist only if the left-hand limit (LHL) and the right-hand limit (RHL) are equal. Because the greatest-integer function $$[x]$$ behaves differently on $$(-1,0)$$ and on $$(0,1)$$, we examine the two sides separately.

1. Preliminary expansions valid for both sides

The standard Maclaurin expansions are

$$\sin t=t-\dfrac{t^{3}}{6}+O(t^{5}),\qquad \tan y=y+\dfrac{y^{3}}{3}+O(y^{5}).$$

For small $$x$$ we first expand $$\sin^{2}x$$:

$$\sin^{2}x=(x-\tfrac{x^{3}}{6}+O(x^{5}))^{2}=x^{2}-\dfrac{x^{4}}{3}+O(x^{6}).$$

Hence

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

Putting this into the expansion of $$\tan$$ we obtain

$$\tan(\pi\sin^{2}x)=\Bigl(\pi x^{2}-\dfrac{\pi x^{4}}{3}+O(x^{6})\Bigr) +\dfrac{1}{3}\Bigl(\pi x^{2}\Bigr)^{3}+O(x^{8}) =\pi x^{2}+O(x^{4}).$$

Thus up to the order required for division by $$x^{2}$$,

$$\tan(\pi\sin^{2}x)=\pi x^{2}+o(x^{2}). \quad -(1)$$

2. Right-hand limit ($$x\to 0^{+}$$)

When $$0\lt x\lt 1$$ we have $$[x]=0.$$ Therefore

$$\sin(x[x])=\sin(x\cdot 0)=\sin 0=0,$$

and, since $$x\gt 0$$, $$|x|=x.$$ So the second part of the numerator becomes

$$\bigl(|x|-\sin(x[x])\bigr)^{2}=(x-0)^{2}=x^{2}. \quad -(2)$$

Adding (1) and (2) we get the entire numerator:

$$\tan(\pi\sin^{2}x)+\bigl(|x|-\sin(x[x])\bigr)^{2} =\bigl(\pi x^{2}+o(x^{2})\bigr)+x^{2} =(\pi+1)x^{2}+o(x^{2}).$$

Dividing by $$x^{2}$$ gives the right-hand limit

$$\text{RHL}=\pi+1.$$

3. Left-hand limit ($$x\to 0^{-}$$)

For $$-1\lt x\lt 0$$ we have $$[x]=-1.$$ Because $$x$$ is negative, $$|x|=-x.$$ Also,

$$\sin(x[x])=\sin\bigl(x\cdot(-1)\bigr)=\sin(-x)=-\sin x.$$

Hence

$$|x|-\sin(x[x])=-x-(-\sin x)=-x+\sin x. \quad -(3)$$

Using $$\sin x=x-\dfrac{x^{3}}{6}+O(x^{5})$$ we obtain

$$-x+\sin x=-x+\Bigl(x-\dfrac{x^{3}}{6}+O(x^{5})\Bigr) =-\dfrac{x^{3}}{6}+O(x^{5}).$$

Squaring,

$$\bigl(|x|-\sin(x[x])\bigr)^{2}=\Bigl(-\dfrac{x^{3}}{6}+O(x^{5})\Bigr)^{2} =\dfrac{x^{6}}{36}+O(x^{8}). \quad -(4)$$

This is of order $$x^{6}$$ and hence negligible compared with the $$x^{2}$$ term we shall divide by.

From (1) and (4) the whole numerator on the left side is

$$\tan(\pi\sin^{2}x)+\bigl(|x|-\sin(x[x])\bigr)^{2} =\pi x^{2}+o(x^{2}).$$

Therefore the left-hand limit is

$$\text{LHL}=\pi.$$

4. Comparison of the two sides

$$\text{RHL}=\pi+1,\qquad \text{LHL}=\pi.$$

Since $$\text{RHL}\neq\text{LHL},$$ the limit as $$x\to 0$$ does not exist.

Hence, the correct answer is Option A.