Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>Let \(f(x) = \frac{\log_e(1+ax) - \log_e(1-bx)}{x}, x \neq 0\). The value to be assigned to \(f(x)\) at \(x = 0\) so that \(f(x)\) is continuous at \(x = 0\) is:</p>
<p>(a) \(a - b\)</p>
<p>(b) \(a + b\)</p>
<p>(c) \(\log_e(ab)\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: For continuity at x=0, we need f(0) = lim(x→0) f(x). Use the standard limit formula: lim(x→0) [log_e(1+u)]/u = 1 to evaluate the indeterminate form.
<p><strong>Step 1:</strong> Recognize that direct substitution gives 0/0 form (indeterminate).</p><p><strong>Step 2:</strong> Rewrite using logarithm properties:</p><p>f(x) = [log_e(1+ax) - log_e(1-bx)]/x = [log_e((1+ax)/(1-bx))]/x</p><p><strong>Step 3:</strong> Expand using series or rewrite as:</p><p>lim(x→0) [log_e(1+ax) - log_e(1-bx)]/x = lim(x→0) [log_e(1+ax)/x - log_e(1-bx)/x]</p><p><strong>Step 4:</strong> Apply standard limit lim(u→0) log_e(1+u)/u = 1:</p><p>= lim(x→0) [a·log_e(1+ax)/(ax) - (-b)·log_e(1-bx)/(-bx)]</p><p>= a·(1) + b·(1) = a + b</p><p><strong>Step 5:</strong> For continuity at x=0, assign f(0) = a + b</p><p>∴ Answer: B</p>
Correct Answer: B