<p>If \(\lim_{x \to 0} \dfrac{\log(3+x) - \log(3-x)}{x} = k\), the value of \(k\) is</p>
Step-by-Step Solution
Key Concept: Use logarithm properties to combine the logs, then recognize this limit as the derivative of log(3+x) at x=0, or apply L'Hôpital's rule to the 0/0 indeterminate form.
<p><strong>Step 1:</strong> Recognize the indeterminate form. As x→0: numerator → log(3) - log(3) = 0 and denominator → 0, giving 0/0 form.</p><p><strong>Step 2:</strong> Simplify using logarithm properties: log(3+x) - log(3-x) = log[(3+x)/(3-x)]</p><p><strong>Step 3:</strong> Rewrite the limit: lim(x→0) [log((3+x)/(3-x))]/x</p><p><strong>Step 4:</strong> Apply L'Hôpital's Rule (or recognize as derivative). Differentiating numerator and denominator:</p><p>Numerator: d/dx[log((3+x)/(3-x))] = 1/[(3+x)/(3-x)] · d/dx[(3+x)/(3-x)]</p><p>d/dx[(3+x)/(3-x)] = [(3-x)(1) - (3+x)(-1)]/(3-x)² = [3-x+3+x]/(3-x)² = 6/(3-x)²</p><p>So numerator derivative = (3-x)/(3+x) · 6/(3-x)² = 6/[(3+x)(3-x)]</p><p>Denominator derivative: 1</p><p><strong>Step 5:</strong> Evaluate at x=0: k = 6/[(3)(3)] = 6/9 = <strong>2/3</strong></p><p>∴ Answer: C (k = 2/3)</p>
Correct Answer: C