Question:

Step 1: Analyze the behavior of the greatest integer function near $x = 0$. For $x$ in a small neighborhood of $0$, specifically when $x \in (-1, 0) \cup (0, 1)$, we have $[x] = 0$. Therefore: $$\sin(x[x]) = \sin(x \cdot 0) = \sin(0) = 0$$ Step 2: Simplify the original expression using the result from Step 1. Since $\sin(x[x]) = 0$, the term $(|x| - \sin(x[x]))^2$ becomes $(|x| - 0)^2 = |x|^2 = x^2$. The limit now simplifies to: $$\lim_{x \to 0} \frac{\tan(\pi \sin^2 x) + x^2}{x^2}$$ Step 3: Split the limit into two separate terms. We can rewrite the limit as: $$\lim_{x \to 0} \frac{\tan(\pi \sin^2 x)}{x^2} + \lim_{x \to 0} \frac{x^2}{x^2}$$ Step 4: Evaluate the first term using Taylor series approximations. As $x \to 0$, we use the approximation $\sin x \approx x$, which gives us: $$\sin^2 x \approx x^2$$ Therefore: $$\tan(\pi \sin^2 x) \approx \tan(\pi x^2)$$ For small arguments, $\tan(u) \approx u$, so: $$\tan(\pi x^2) \approx \pi x^2$$ Thus: $$\lim_{x \to 0} \frac{\tan(\pi \sin^2 x)}{x^2} = \lim_{x \to 0} \frac{\pi x^2}{x^2} = \pi$$ Step 5: Evaluate the second term. $$\lim_{x \to 0} \frac{x^2}{x^2} = 1$$ Step 6: Combine both terms to find the final answer. $$\lim_{x \to 0} \frac{\tan(\pi \sin^2 x) + (|x| - \sin(x[x]))^2}{x^2} = \pi + 1$$ The answer is $\boxed{\pi + 1}$, which corresponds to **Option 2**.