Let f(x) = <span class="math-inline">\( \frac{-x(1+|1-|x||)}{|1-x|} \)</span> for <span class="math-inline">\( x \neq 1 \)</span>. Then <span class="math-inline">\( \lim_{x \to 1^-} f(x) \)</span> does not exist.
lim<sub>x \to 1^-</sub> f(x) does not exist
lim<sub>x \to 1^-</sub> f(x) does not exist
lim<sub>x \to 1^-</sub> f(x) = 0
lim<sub>x \to 1^-</sub> f(x) = 0
Step-by-Step Solution
Key Concept: When evaluating limits as x approaches 1 from the left, we must carefully simplify nested absolute values by considering the sign of expressions in the domain x < 1. The key is to determine whether the left-hand limit exists and is finite.
<p><strong>Step 1: Simplify for x → 1⁻ (where 0 < x < 1)</strong></p><p>For x < 1, we have 1 - x > 0, so |1 - x| = 1 - x.</p><p>Also, for 0 < x < 1, we have |x| = x, so 1 - |x| = 1 - x > 0.</p><p>Therefore |1 - |x|| = 1 - x.</p><p><strong>Step 2: Substitute into the function</strong></p><p>$$f(x) = \frac{-x(1 + |1 - |x||)}{|1 - x|} = \frac{-x(1 + (1-x))}{1-x} = \frac{-x(2-x)}{1-x}$$</p><p><strong>Step 3: Evaluate the left-hand limit</strong></p><p>$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{-x(2-x)}{1-x}$$</p><p>As x → 1⁻: numerator → -1(2-1) = -1 and denominator → 0⁺</p><p>$$\lim_{x \to 1^-} \frac{-x(2-x)}{1-x} = \frac{-1}{0^+} = -\infty$$</p><p><strong>Step 4: Check behavior from the right (for completeness)</strong></p><p>For 1 < x < 2: |x| = x, 1 - |x| = 1 - x < 0, so |1 - |x|| = x - 1.</p><p>Also |1 - x| = x - 1 (since x > 1).</p><p>$$f(x) = \frac{-x(1 + (x-1))}{x-1} = \frac{-x \cdot x}{x-1} = \frac{-x^2}{x-1}$$</p><p>As x → 1⁺: $$\lim_{x \to 1^+} \frac{-x^2}{x-1} = \frac{-1}{0^+} = -\infty$$</p><p><strong>Step 5: Conclusion</strong></p><p>Although both one-sided limits approach -∞ (same direction), the limit is typically said to "not exist" in the finite sense since -∞ is not a real number. The standard interpretation is that the limit does not exist as a finite value.</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A