Question:

<p><strong>Step 1: Analyze the expression structure</strong></p><p>The limit is: $\lim_{x \to 0} \frac{\tan(\pi \sin^2 x) + (|x| - \sin(x[x]))^2}{x^2}$</p><p>We must examine left and right limits separately due to the floor function [x] and absolute value |x|.</p><p><strong>Step 2: Right-hand limit (x → 0⁺)</strong></p><p>For x ∈ (0, 1): [x] = 0, so x[x] = 0</p><p>Also |x| = x for x > 0</p><p>Numerator: $\tan(\pi \sin^2 x) + (x - \sin(0))^2 = \tan(\pi \sin^2 x) + x^2$</p><p>Using $\sin x \approx x$ for small x: $\sin^2 x \approx x^2$</p><p>So $\tan(\pi x^2) \approx \pi x^2$ (since $\tan u \approx u$ for small u)</p><p>Therefore: $\lim_{x \to 0^+} \frac{\pi x^2 + x^2}{x^2} = \lim_{x \to 0^+} \frac{(\pi + 1)x^2}{x^2} = \pi + 1$</p><p><strong>Step 3: Left-hand limit (x → 0⁻)</strong></p><p>For x ∈ [-1, 0): [x] = -1, so x[x] = -x</p><p>Also |x| = -x for x < 0</p><p>Numerator: $\tan(\pi \sin^2 x) + (-x - \sin(-x))^2$</p><p>Since $\sin(-x) = -\sin(x)$:</p><p>$(-x - (-\sin x))^2 = (-x + \sin x)^2 = (\sin x - x)^2$</p><p>Using $\sin x \approx x - \frac{x^3}{6}$:</p><p>$(\sin x - x)^2 \approx (-\frac{x^3}{6})^2 = O(x^6)$ (negligible compared to $x^2$)</p><p>Therefore: $\lim_{x \to 0^-} \frac{\pi x^2 + O(x^6)}{x^2} = \pi$</p><p><strong>Step 4: Compare the one-sided limits</strong></p><p>Left-hand limit: $\pi$</p><p>Right-hand limit: $\pi + 1$</p><p>Since $\pi \neq \pi + 1$, the left and right limits are not equal.</p><p><strong>∴ Answer:</strong> The limit does not exist. The answer is 4.</p>