For each <span class="math-inline">\( t \in \mathbb{R} \)</span>, let <span class="math-inline">\( [t] \)</span> be the greatest integer less than or equal to t. Then <span class="math-inline">\( \lim_{x \to 1^-} \frac{(1-|x|+\sin[1-x])\sin\left(\frac{\pi}{2}[1-x]\right)}{|1-x|[|1-x|]} \)</span>
Step-by-Step Solution
Key Concept: As x → 1⁻, we have x < 1, so |x| = x and 1-x > 0. We must analyze the floor function [1-x] which takes discrete values as x approaches 1 from the left, making the limit behavior discontinuous.
<p><strong>Step 1: Analyze the domain as x → 1⁻</strong></p><p>As x → 1⁻, we have x < 1, so:</p><ul><li>|x| = x</li><li>1 - x → 0⁺</li><li>|1-x| = 1-x</li></ul><p><strong>Step 2: Determine the floor function value</strong></p><p>For x sufficiently close to 1⁻ (say 0 < 1-x < 1), we have [1-x] = 0.</p><p><strong>Step 3: Evaluate the numerator</strong></p><p>Numerator = (1 - |x| + sin[1-x]) · sin(π/2 · [1-x])</p><p>= (1 - x + sin(0)) · sin(π/2 · 0)</p><p>= (1 - x + 0) · sin(0)</p><p>= (1 - x) · 0 = 0</p><p><strong>Step 4: Analyze the denominator</strong></p><p>Denominator = |1-x| · [|1-x|] = (1-x) · [1-x]</p><p>For 0 < 1-x < 1, we have [1-x] = 0</p><p>So Denominator = (1-x) · 0 = 0</p><p><strong>Step 5: Recognize the indeterminate form and its pathology</strong></p><p>We get 0/0 form, but the denominator is identically 0 for all x in a left-neighborhood of 1 (where 0 < 1-x < 1), while the numerator is also 0. The expression becomes 0/0 consistently, making the function undefined in a deleted neighborhood of 1.</p><p>For any sequence xₙ → 1⁻, the expression is of the form 0/0 with no well-defined limiting behavior. Attempting to evaluate from different directions or using L'Hôpital's rule fails due to the floor function's discontinuous nature.</p><p><strong>∴ Answer: The limit does not exist (Option 3)</strong></p>
Correct Answer: 3